Reorganize & alg-1

5 files changed, 1000 insertions, 0 deletions

diff --git a/calc/sol0113.tex b/calc/sol0113.tex
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+++ b/calc/sol0113.tex
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+\documentclass[10pt,a5paper]{article}
+\usepackage[svgnames, rgb]{xcolor}
+
+\input{intro}
+
+\lhead{\color{gray} Шарафатдинов Камиль 192}
+\rhead{\color{gray} ДЗ к 27.01 (\texttt{sol0113 + sol0120})}
+\title{ДЗ на 27.01}
+\author{Шарафатдинов Камиль БПМИ-192}
+\date{билд: \today}
+
+
+% -- Here bet  dragons --
+\begin{document}\thispagestyle{empty}
+
+\maketitle
+\clearpage
+\setcounter{page}{1}
+\[
+    \abs{\frac{1}{2}}
+\]
+\end{document}
diff --git a/calc/sol0120.tex b/calc/sol0120.tex
new file mode 100644
index 0000000..f073973
--- /dev/null
+++ b/calc/sol0120.tex
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+\documentclass[10pt,a5paper]{article}
+\usepackage[svgnames, rgb]{xcolor}
+
+\input{intro}
+
+\lhead{\color{gray} Шарафатдинов Камиль 192}
+\rhead{\color{gray} ДЗ к 27.01 (\texttt{sol0113 + sol0120})}
+\title{ДЗ на 27.01}
+\author{Шарафатдинов Камиль БПМИ-192}
+\date{билд: \today}
+
+
+% -- Here bet  dragons --
+\begin{document}\thispagestyle{empty}
+
+\maketitle
+\clearpage
+\setcounter{page}{1}
+
+\question{Лемма 1}{
+    \[
+        \int \frac{dx}{x^s} = \frac{1}{(1 - s)x^{s - 1}} + C
+    \]
+}
+    \[
+        \ br{ \frac{1}{(1 - s)x^{s - 1}} + C }' =
+        -\frac{0 - (1 - s)(s - 2)x^{s - 2}}{(1 - s)^2 x^{2s - 2}} =
+        \frac{1}{x^s} \qed
+    \]
+
+\question{Лемма 2}{
+    \[
+        \int \frac{dx}{(x^2 + a^2)^2} = \frac{1}{2a^2} \ br{
+            \frac{x}{x^2 + a^2} + \frac{1}{a}\arctan\frac{x}{a} 
+        } + C
+    \]
+}
+
+    \begin{align*}
+        \ br{
+            \frac{1}{2a^2} \ br{
+                \frac{x}{x^2 + a^2} + \frac{1}{a}\arctan\frac{x}{a}
+            }
+        }' =
+        \frac{1}{2a^2} \ br{
+            \frac{x^2 + a^2 - 2x^2}{(x^2 + a^2)^2} + \frac{1}{x^2 + a^2}
+        } =
+        \frac{1}{2a^2} \ br{
+            \frac{2a^2}{(x^2 + a^2)^2}
+        } =
+        \frac{1}{(x^2 + a^2)^2} \qed
+    \end{align*}
+
+\question{(seminar0113) 7.3}{
+    \[
+        \int \frac{dx}{x^4 + 4} = \frac{
+            \log | x^2 + 2x + 2 | + 2\arctan(x + 1) -
+            \log | x^2 - 2x + 2 | + 2\arctan(x - 1)
+        }{16} + \bar{C}
+    \]
+}
+
+    \[
+        x^4 + 4 = (x - (1 + i))(x - (i - 1))(x - (-i - 1))(x - (-i + 1)) =
+        (x^2 + 2x + 2)(x^2 - 2x + 2)
+    \]
+
+    \[
+        \frac{1}{x^4 + 4} = \frac{Ax + B}{x^2 + 2x + 2} + \frac{Cx + D}{x^2 - 2x + 2}
+    \]
+
+    \[
+        (Ax + B)(x^2 - 2x + 2) + (Cx + D)(x^2 + 2x + 2) \equiv 1
+    \]
+
+    С помощью давно забытой китайской техники решения систем уравнений получаем:
+    \[\begin{cases*}
+        A = \frac{1}{8}\\
+        B = \frac{1}{4}\\
+        C = -\frac{1}{8}\\
+        D = \frac{1}{4}\\
+    \end{cases*}\]
+
+    \[
+        \int \frac{1}{x^4 + 4} =
+            \frac{1}{8}\int \frac{x + 2}{x^2 + 2x + 2} dx +
+            \frac{1}{8}\int \frac{2 - x}{x^2 - 2x + 2} dx
+    \]
+
+    \begin{minipage}{0.45\textwidth}
+        \setlength{\jot}{16pt}
+        \begin{gather*}
+            \int \frac{x + 2}{x^2 + 2x + 2} dx =\\
+            \int \frac{x + 2}{(x + 1)^2 + 1} dx =\\
+            \int \frac{(x + 1) dx}{(x + 1)^2 + 1} + \int \frac{dx}{(x + 1)^2 + 1} =\\
+            =\begin{bmatrix} \frac{d(x^2 + 2x + 2)}{2} = (x + 1)dx \end{bmatrix} =\\
+            \int \frac{\frac{1}{2}d( (x + 1)^2 + 1 )}{(x + 1)^2 + 1} + \arctan(x + 1) =\\
+            = \frac{1}{2}\log | x^2 + 2x + 2 | + \arctan(x + 1) + C_1
+        \end{gather*}
+    \end{minipage}
+    \begin{minipage}{0.45\textwidth}
+        \begin{tabular}{|p{\textwidth}}
+        \setlength{\jot}{16pt}
+        \begin{gather*}
+            \int \frac{2 - x}{x^2 - 2x + 2} dx =\\
+            -\int \frac{x - 2}{(x - 1)^2 + 1} dx =\\
+            -\int \frac{(x - 1) dx}{(x - 1)^2 + 1} + \int \frac{dx}{(x - 1)^2 + 1} =\\
+            =\begin{bmatrix} \frac{d(x^2 - 2x + 2)}{2} = (x - 1)dx \end{bmatrix} =\\
+            -\int \frac{\frac{1}{2}d( (x - 1)^2 + 1 )}{(x - 1)^2 + 1} + \arctan(x - 1) =\\
+            = -\frac{1}{2}\log | x^2 - 2x + 2 | + \arctan(x - 1) + C_2
+        \end{gather*}
+        \end{tabular}
+    \end{minipage}    
+
+    \begin{gather*}
+        \frac{1}{8}\int \frac{x + 2}{x^2 + 2x + 2} dx +
+        \frac{1}{8}\int \frac{2 - x}{x^2 - 2x + 2} dx =\\
+        =\frac{1}{16}\ br{
+            \log | x^2 + 2x + 2 | + 2\arctan(x + 1) -
+            \log | x^2 - 2x + 2 | + 2\arctan(x - 1)
+        } + \bar{C}
+    \end{gather*}
+
+\question{(seminar0113) 8.b}{
+    \[
+        \int \frac{x^5 - x}{x^8 + 1}dx = \frac{\sqrt{2}}{8} \ br{
+            \log |x^4 - \sqrt{2} x^2 + 1| - \log |x^4 + \sqrt{2} x^2 + 1|
+        } + \bar{C}
+    \]
+}
+    
+    \[
+        \int \frac{x^5 - x}{x^8 + 1}dx =
+        \int \frac{x(x^4 - 1}{x^8 + 1}dx =
+        \begin{bmatrix}
+        u = x^2\\
+        dx = \frac{du}{2x}    
+        \end{bmatrix} =
+        \int \frac{x(u^2 - 1)}{u^4 + 1}\frac{du}{2x} =
+        \frac{1}{2} \int \frac{u^2 - 1}{u^4 + 1} du
+    \]
+    \[
+        u^4 + 1 = \ br{u^2 + \sqrt{2} u + 1}\ br{u^2 - \sqrt{2} u + 1}
+    \]
+    \[
+        \frac{u^2 - 1}{u^4 + 1} =
+        \frac{Au + B}{u^2 + \sqrt{2} u + 1} +
+        \frac{Cu + D}{u^2 - \sqrt{2} u + 1}
+    \]
+
+    \[
+        (Au + B)(u^2 - \sqrt{2}u + 1) + (Cu + D)(u^2 + \sqrt{2}u + 1) \equiv u^2 - 1
+    \]
+
+    Все тем же китайским методом:
+    \[\begin{cases*}
+        A = -\frac{\sqrt{2}}{2}\\
+        B = -\frac{1}{2}\\
+        C = \frac{\sqrt{2}}{2}\\
+        D = -\frac{1}{2}\\
+    \end{cases*}\]
+
+    \newcommand{\invsq}{\frac{\sqrt{2}}{2}}
+    \[
+        \int \frac{u^2 - 1}{u^4 + 1} du =
+        \invsq \int \frac{-u - \invsq}{u^2 + \sqrt{2} u + 1} du +
+        \invsq \int \frac{u - \invsq}{u^2 - \sqrt{2} u + 1} du
+    \]    
+    
+    \begin{minipage}{0.45\textwidth}
+        \setlength{\jot}{16pt}
+        \begin{gather*}
+            \int \frac{-u - \invsq}{u^2 + \sqrt{2} u + 1} du =\\
+            -\int \frac{u + \invsq}{u^2 + \sqrt{2} u + 1} du =\\
+            -\frac{1}{2} \int \frac{d \ br{ u^2 + \sqrt{2} u + 1 }}{u^2 + \sqrt{2} u + 1} =\\
+            -\frac{1}{2} \log |u^2 + \sqrt{2} u + 1| + C_1
+        \end{gather*}
+    \end{minipage}
+    \begin{minipage}{0.45\textwidth}
+        \begin{tabular}{|p{\textwidth}}
+        \setlength{\jot}{16pt}
+        \begin{gather*}
+            \int \frac{u - \invsq}{u^2 - \sqrt{2} u + 1} du =\\
+            \int \frac{u + \invsq}{u^2 - \sqrt{2} u + 1} du =\\
+            \frac{1}{2} \int \frac{d \ br{ u^2 - \sqrt{2} u + 1 }}{u^2 - \sqrt{2} u + 1} =\\
+            \frac{1}{2} \log |u^2 - \sqrt{2} u + 1| + C_2
+        \end{gather*}
+        \end{tabular}
+    \end{minipage}
+
+    \[
+        \frac{1}{2} \int \frac{u^2 - 1}{u^4 + 1} du =
+        \frac{\sqrt{2}}{8} \ br{
+            \log |u^2 - \sqrt{2} u + 1| - \log |u^2 + \sqrt{2} u + 1|
+        } + C_3
+    \]
+
+    Обратно к $x$:
+    \[
+        \int \frac{x^5 - x}{x^8 + 1}dx =
+        \frac{\sqrt{2}}{8} \ br{
+            \log |x^4 - \sqrt{2} x^2 + 1| - \log |x^4 + \sqrt{2} x^2 + 1|
+        } + \bar{C}
+    \]
+
+\clearpage
+\question{(seminar0113) 13}{
+    \[
+        \int \frac{x \ dx}{(x^2 + 1)(x + 2)(x + 3)} = \frac{1}{20} \ br{
+            -8 \log |x + 2| + 6 \log |x + 3| + \log (x^2 + 1) + 2\arctan x
+        } + \bar{C}
+    \]
+}
+    \[
+        \frac{x}{(x^2 + 1)(x + 2)(x + 3)} =
+        \frac{Ax + B}{x^2 + 1} + \frac{C}{x + 2} + \frac{D}{x + 3}
+    \]
+
+    \[
+        (Ax + B)(x + 2)(x + 3) + C(x^2 + 1)(x + 3) + D(x^2 + 1)(x + 2) \equiv x
+    \]
+    \[\begin{cases*}
+        A = 0.1\\
+        B = 0.1\\
+        C = -0.4\\
+        D = 0.3\\
+    \end{cases*}\]
+
+    \begin{gather*}
+        \int -\frac{2}{5} \frac{dx}{x + 2} = -\frac{2 \log |x + 2|}{5} + C_1\\[16pt]
+        \int \frac{3}{10} \frac{dx}{x + 3} = \frac{3 \log |x + 3|}{10} + C_2\\[16pt]
+        \int \frac{1}{10} \frac{(x + 1) dx}{x^2 + 1} =
+            \frac{1}{10} \ br{
+                \frac{1}{2}\int \frac{2x \ dx}{x^2 + 1} + \int \frac{dx}{x^2 + 1}
+            } =
+            \frac{1}{10} \ br{
+                \frac{1}{2} \log (x^2 + 1) + \arctan(x)
+            } + C_3
+    \end{gather*}
+
+    \[
+        \int \frac{x \ dx}{(x^2 + 1)(x + 2)(x + 3)} = \frac{1}{20} \ br{
+            -8 \log |x + 2| + 6 \log |x + 3| + \log (x^2 + 1) + 2\arctan x
+        } + \bar{C}
+    \]
+
+\question{(seminar0120) 2.4}{
+    \[
+        \int \frac{dx}{x(x^2 + 1)^2} = -\frac{1}{2} \ br{
+            -\log (x^2 + 1) +
+            \frac{1}{x^2 + 1} +
+            2\log |x|
+        } + \bar{C}
+    \]
+}
+
+    \[
+        \frac{1}{x(x^2 + 1)^2} =
+        \frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{(x^2 + 1)^2} + \frac{E}{x}
+    \]
+
+    \[
+        (Ax + B)(x^2 + 1)x + (Cx + D)x + E(x^2 + 1)^2 \equiv 1
+    \]
+
+    \[\begin{cases*}
+        A = -1\\
+        B = 0\\
+        C = -1\\
+        D = 0\\
+        E = 1
+    \end{cases*}\]
+
+    \begin{gather*}
+        \int - \frac{x \ dx}{x^2 + 1} =
+            -\frac{1}{2} \int \frac{2x \ dx}{x^2 + 1} =
+            -\frac{1}{2} \log (x^2 + 1) + C_1\\[12pt]
+        \int - \frac{x \ dx}{(x^2 + 1)^2} =
+            -\frac{1}{2} \int \frac{2x \ dx}{(x^2 + 1)^2} =
+            \begin{bmatrix}
+                \displaystyle \int \frac{dx}{x^s} = \frac{1}{(1 - s)x^{s - 1}} + C
+            \end{bmatrix} =
+            \frac{1}{2(x^2 + 1)} + C_2\\[12pt]
+        \int \frac{dx}{x} = \log |x| + C_3
+    \end{gather*}
+
+    \[
+        \int \frac{dx}{x(x^2 + 1)^2} =
+            -\frac{1}{2} \log (x^2 + 1) +
+            \frac{1}{2(x^2 + 1)} +
+            \log |x| + \bar{C}
+    \]
+
+\question{(seminar0120) 11}{
+    \[
+        \int \frac{dx}{(x^3 + 1)^2}
+    \]
+}
+
+    \[
+        \frac{1}{(x^3 + 1)^2} = \frac{1}{(x + 1)^2(x^2 - x + 1)^2} =
+        \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{Cx + D}{x^2 - x + 1} + \frac{Ex + F}{(x^2 - x + 1)^2}
+    \]
+
+    \[
+        A(x^2 - x + 1)^2(x + 1) +
+        B(x^2 - x + 1)^2 +
+        (Cx + D)(x + 1)^2(x^2 - x + 1) +
+        (Ex + F)(x + 1)^2 \equiv 1
+    \]
+    \[\begin{cases*}
+        A = 2/9\\
+        B = 1/9\\
+        C = -2/9, \ \ 
+        D = 1/3\\
+        E = -1/3, \ \ 
+        F = 1/3\\
+    \end{cases*}\]
+
+    \begin{align}
+        \int \frac{2dx}{9(x + 1)} &= \frac{2}{9} \log |x + 1| + C_1 &\\[8pt]
+        \int \frac{dx}{9(x + 1)^2} & = -\frac{1}{9(x + 1)} + C_2 &
+            \begin{bmatrix}
+                \text{Лемма 1}
+            \end{bmatrix}\\[8pt]
+        \int \frac{-2x + 3}{9(x^2 - x + 1)}dx &=
+            -\frac{1}{9} \ br{
+                \int \frac{(2x - 1) dx}{x^2 - x + 1} -
+                \int \frac{2 dx}{\ br{ x - \frac{1}{2} }^2 + \frac{3}{4}}
+            }\nonumber \\[8pt]
+            &=
+            -\frac{1}{9} \ br{
+                \log (x^2 - x + 1) -
+                \frac{4}{\sqrt{3}} \arctan \frac{2x - 1}{\sqrt{3}}
+            } + C_3\\[8pt]
+        \frac{1}{3} \int \frac{(1 - x)dx}{(x^2 - x + 1)^2} &=
+            -\frac{1}{6} \ br{
+                \int \frac{(2x - 1)dx}{(x^2 - x + 1)^2} -
+                \int \frac{dx}{\ br{ \ br{ x - \frac{1}{2} }^2 + \frac{3}{4} }^2}
+            }\nonumber \\[8pt]
+            &=
+            -\frac{1}{6} \ br{
+                -\frac{1}{x^2 - x + 1} +
+                \frac{2}{3} \ br{
+                    \frac{x}{x^2 - x + 1} +
+                    \frac{2}{\sqrt{3}} \arctan\frac{2x - 1}{\sqrt{3}}
+                }
+            } + C_4
+            &\begin{bmatrix}
+                \text{Лемма 1 на левую часть}\\
+                \text{Лемма 2 на правую часть}
+            \end{bmatrix}
+            \nonumber \\[8pt]
+            &= \frac{1}{9} \ br{
+                \frac{2x - 1}{x^2 - x + 1} -
+                \frac{2}{\sqrt{3}} \arctan \frac{2x - 1}{\sqrt{3}}
+            } + C_4
+    \end{align}
+
+    \begin{gather*}
+        \int \frac{1}{(x^3 + 1)^2} =
+        \int \frac{2dx}{9(x + 1)} +
+        \int \frac{dx}{9(x + 1)^2} +
+        \int \frac{-2x + 3}{9(x^2 - x + 1)}dx +
+        \frac{1}{3} \int \frac{(1 - x)dx}{(x^2 - x + 1)^2} =\\[16pt]
+        \frac{1}{9} \ br{
+            2\log |x + 1| - \frac{1}{x + 1}
+            - \log(x^2 - x + 1) + \frac{4}{\sqrt{3}}\arctan \frac{2x - 1}{\sqrt{3}}
+            - \frac{2x - 1}{x^2 - x + 1} - \frac{2}{\sqrt{3}}\arctan\frac{2x - 1}{\sqrt{3}}
+        } + C
+    \end{gather*}
+\end{document} 
\ No newline at end of file
diff --git a/calc/sol0127.tex b/calc/sol0127.tex
new file mode 100644
index 0000000..1098687
--- /dev/null
+++ b/calc/sol0127.tex
@@ -0,0 +1,268 @@
+\documentclass[10pt,a5paper]{article}
+\usepackage[svgnames, rgb]{xcolor}
+
+\input{intro}
+
+\lhead{\color{gray} Шарафатдинов Камиль 192}
+\rhead{\color{gray} \texttt{sol0127}}
+\title{ДЗ на 03.02}
+\author{Шарафатдинов Камиль БПМИ-192}
+\date{билд: \today}
+
+
+% -- Here bet  dragons --
+\begin{document}\thispagestyle{empty}
+
+\maketitle
+\clearpage
+\setcounter{page}{1}
+
+%\question{8.a}{
+%    \[
+%        \int \frac{2\sin^3 x + \cos^2 x \sin 2x}{\sin^4 x + 3 \cos^4 x} \dif x = \todo + C
+%    \]
+%}
+
+    
+
+\question{8.c}{
+    \[
+        \int \frac{\dif x}{\cosh^3 x + 3\cosh x} = \frac{
+            2\arctan \sinh x - \arctan \frac{\sinh x}{2}
+        }{6} + C
+    \]
+}
+    \begin{align*}
+        \int \frac{\dif x}{\cosh^3 x + 3\cosh x}
+        &= \int \frac{\frac{\dif \sinh x}{\cosh x}}{\cosh^3 x + 3\cosh x}
+            &\explain{
+                \dif \sinh x = \cosh x \dif x
+            } \\[8pt]
+        &= \int \frac{\dif \sinh x}{\cosh^4 x + 3\cosh^2 x} \\[8pt]
+        &= \int \frac{\dif \sinh x}{\br{ 1 + \sinh^2 x }^2 + 3\br {1 + \sinh^2 x}}
+            &\explain{
+                \cosh^2 x - \sinh^2 x = 1
+            } \\[8pt]
+        &= \int \frac{\dif u}{\br{ 1 + u^2 }^2 + 3\br {1 + u^2}}
+            &\explain{
+                u = \sinh x
+            } \\[8pt]
+        &= \int \frac{\dif u}{4 + 5u^2 + u^4} \\[8pt]
+        &= \int \frac{\dif u}{\br{ u^2 + 1 } \br{ u^2 + 4 }} \\[8pt]
+        &= \int \frac{\dif u}{3} \br{
+            \frac{1}{ u^2 + 1 } - \frac{1}{ u^2 + 4 }
+        } \\[8pt]
+        &= \frac{1}{3} \br{
+            \int \frac{du}{u^2 + 1} - \int \frac{du}{u^2 + 4}
+        } \\[8pt]
+        &= \frac{1}{3} \br{
+            \arctan u - \frac{1}{2}\arctan \frac{u}{2}
+        } + C
+            &\explain{
+                u = \sinh x
+            }\\[8pt]
+        &= \frac{1}{6} \br{
+            2\arctan \sinh x - \arctan \frac{\sinh x}{2}
+        } + C
+    \end{align*}
+
+\clearpage
+\question{8.e}{
+    \[
+        \int \frac{\dif x}{\sin^4 x + \cos^4 x} = \frac{\sqrt{2}}{2} \br{
+            \arctan \br{ \sqrt{2} \tan x + 1 } + \arctan  \br{ \sqrt{2} \tan x - 1 }
+        } + C
+    \]
+}
+
+    \[
+        \sin^4 x + \cos^4 x =
+        \br{ 1 - \cos^2 x }^2 + \cos^4 x =
+        1 - 2\cos^2 x + 2\cos^4 x
+    \]
+    \[
+        \dif x = \cos^2 x \dif\ (\tan x)
+    \]
+    \[
+        1 + \tan^2 x = \frac{1}{\cos^2 x}
+    \]
+
+    \begin{align*}
+        \int \frac{\dif x}{\sin^4 x + \cos^4 x}
+            &= \int \frac{\cos^2 x \dif\ (\tan x)}{1 - 2\cos^2 x + 2\cos^4 x} \\[8pt]
+            &= \int \frac{\dif\ (\tan x)}{\frac{1}{\cos^2 x} - 2 + 2\cos^2 x} \\[8pt]
+            &= \int \frac{\dif\ (\tan x)}{1 + \tan^2 x - 2 + \frac{2}{1 + \tan^2 x}} \\[8pt]
+            &= \int \frac{\dif u}{-1 + u^2 + \frac{2}{1 + u^2}} & [u = \tan x]\\[8pt]
+            &= \int \frac{(1 + u^2) \dif u}{u^4 + 1} \\[8pt]
+    \end{align*}
+
+    По прошлой домашке мы знаем, что
+    \[
+        x^4 + 1 = (x^2 - \sqrt{2} x + 1)(x^2 + \sqrt{2} x + 1)
+    \]
+
+    Разложим на слагаемые:
+    \[
+        \frac{x^2 + 1}{x^4 + 1} =
+        \frac{1}{2} \br{ \frac{1}{x^2 + \sqrt{2} x + 1} + \frac{1}{x^2 - \sqrt{2} + 1} }
+    \]
+
+    \begin{align*}
+        \int \frac{1}{u^2 + \sqrt{2} u + 1}\dif u
+            &= \int \frac{1}{\br{ u + \frac{\sqrt{2}}{2} }^2 + \frac{1}{2}}\dif u \\[8pt]
+            &= \sqrt{2} \arctan \br{ \sqrt{2} u + 1 } + C_1 \\[16pt]
+        \int \frac{1}{u^2 - \sqrt{2} u + 1}\dif u
+            &= \int \frac{1}{\br{ u - \frac{\sqrt{2}}{2} }^2 + \frac{1}{2}}\dif u \\[8pt]
+            &= \sqrt{2} \arctan \br{ \sqrt{2} u - 1 } + C_2 \\[8pt]
+    \end{align*}
+
+    \begin{gather*}
+        \int \frac{\dif x}{\sin^4 x + \cos^4 x} =
+        \int \frac{(\tan^2 x + 1) \dif\ \tan x}{\tan^4 x + 1} =\\[8pt] =
+        \frac{\sqrt{2}}{2} \br{
+            \arctan \br{ \sqrt{2} \tan x + 1 } + \arctan  \br{ \sqrt{2} \tan x - 1 }
+        } + C
+    \end{gather*}
+
+\question{8.g}{
+    \[
+        \int \frac{\dif x}{a \sin x + b \cos x + c}, \qquad c > \sqrt{a^2 + b^2}
+    \]
+}
+
+    Найдем такой интеграл в предположении $a > 1$:
+
+    \begin{align*}
+        \int \frac{\dif x}{\sin x + a}
+        &= \int \frac{\frac{2 \dif u}{1 + u^2}}{\frac{2u}{1 + u^2} + a}
+            &\explain{
+                \displaystyle u = \tan \frac{x}{2}\\[8pt]
+                \displaystyle \dif x = \frac{2 \dif u}{1 + u^2}\\[8pt]
+                \displaystyle \sin x = \frac{2u}{1 + u^2}
+            } \\[8pt]
+        &= \int \frac{2\dif u}{2u + a + au^2} \\[8pt]
+        &= \int \frac{2\dif u}
+            {\br{ \sqrt{a} u + \frac{1}{\sqrt{a}} }^2 + 1 - \frac{1}{a}} \\[8pt]
+        &= \int \frac{2\dif u}
+            {\br{ \sqrt{a} u + \frac{1}{\sqrt{a}} }^2 + \frac{a^2 - 1}{a}} \\[8pt]
+        &= \frac{2}{\sqrt{a}}\int \frac{\dif v}{v^2 + \frac{a^2 - 1}{a}}
+            &\explain{
+                v = \sqrt{a} u + \frac{1}{\sqrt{a}}\\
+                \dif v = \sqrt{a} \dif u
+            } \\[8pt]
+        &= \frac{2\sqrt{a}}{\sqrt{a}\sqrt{a^2 - 1}}
+            \arctan \br{ \frac{\sqrt{a} v}{\sqrt{a^2 - 1}} } + C \\[8pt]
+        &= \frac{2}{\sqrt{a^2 - 1}}
+            \arctan \br{ \frac{\sqrt{a} v}{\sqrt{a^2 - 1}} } + C \\[8pt]
+        &= \frac{2}{\sqrt{a^2 - 1}}
+            \arctan \br{ \frac{au + 1}{\sqrt{a^2 - 1}} } + C
+            &\explain{
+                v = \sqrt{a} u + \frac{1}{\sqrt{a}}
+            } \\[8pt]
+        &= \frac{2}{\sqrt{a^2 - 1}}
+            \arctan \br{ \frac{a \tan \frac{x}{2} + 1}{\sqrt{a^2 - 1}} } + C
+    \end{align*}
+
+    Теперь, непосредственно задание
+
+    \begin{align*}
+        \int \frac{\dif x}{a \sin x + b \cos x + c}
+        &= \int \frac{\dif x}{r \br{ \frac{a}{r} \sin x + \frac{b}{r} \cos x} + c }
+            &\explain{
+                \displaystyle r = \sqrt{a^2 + b^2}
+            } \\[8pt]
+        &= \frac{1}{r} \int \frac{\dif x}{\cos \phi \sin x + \sin \phi \cos x + c/r}
+            &\explain{
+                \displaystyle \phi = \arccos \frac{a}{r} = \arcsin \frac{b}{r}
+            } \\[8pt]
+        &= \frac{1}{r} \int \frac{\dif x}{\sin \br{ \phi + x } + c/r} \\[8pt]
+        &= \frac{1}{r} \int \frac{\dif u}{\sin \br{ u } + c/r}
+            &\explain{
+                u = \phi + x\\
+                du = dx
+            } \\[8pt]
+        &= \frac{2}{r\sqrt{\dfrac{c^2}{r^2} - 1}}
+            \arctan \br{
+                \frac
+                    {\displaystyle \frac{c}{r} \tan \frac{u}{2} + 1}
+                    {\displaystyle \sqrt{\frac{c^2}{r^2} - 1}}
+            } + C
+            &\explain{\text{По доказанному}}\\[8pt]
+        &= \frac{2}{\sqrt{c^2 - r^2}}
+            \arctan \br{
+                \frac
+                    {\displaystyle \frac{c}{r} \tan \frac{\arccos \dfrac{a}{r}}{2} + 1}
+                    {\displaystyle \sqrt{\frac{c^2}{r^2} - 1}}
+            } + C \\[8pt]
+        &= \frac{2}{\sqrt{c^2 - r^2}}
+            \arctan \br{
+                \frac
+                    {\displaystyle c \tan \frac{\arccos \dfrac{a}{r}}{2} + r}
+                    {\displaystyle \sqrt{c^2 - r^2}}
+            } + C \\[8pt]
+        &= \frac{2}{\sqrt{c^2 - a^2 - b^2}}
+            \arctan \br{
+                \frac
+                    {\displaystyle c \tan \frac{\arccos \dfrac{a}{\sqrt{a^2 + b^2}}}{2} + \sqrt{a^2 + b^2}}
+                    {\displaystyle \sqrt{c^2 - a^2 - b^2}}
+            } + C \\[8pt]
+    \end{align*}
+
+\clearpage
+
+\question{10}{
+    \[
+        \int \frac{\dif x}{(a\sin x + b\cos x)^n}
+    \]
+}
+
+    Найдем рекуррентную формулу для следующего интеграла:
+    \begin{align*}
+        \int \frac{\dif x}{\sin^n x}
+        &= -\int \frac{\dif \cos x}{\sin^{n + 1} x}\\[6pt]
+        &= -\frac{\cos x}{\sin^{n + 1} x} -
+            \int \br{ \frac{1}{\sin^{n + 1} x} }' \cos x \dif x
+            &\explain{
+                \displaystyle \int Fg \dif x = FG - \int fG \dif x\\
+                \displaystyle F = \frac{1}{\sin^{n + 1} x}\\[10pt]
+                \displaystyle g = \frac{\dif \cos x}{\dif x}\\
+                \displaystyle G = \cos x
+            } \\[8pt]
+        &= -\frac{\cos x}{\sin^{n + 1} x} -
+            (n + 1) \int \frac{\cos^2 x \dif x}{\sin^{n + 2} x} \\[8pt]
+        &= -\frac{\cos x}{\sin^{n + 1} x} -
+            (n + 1) \int \frac{(1 - \sin^2 x) \dif x}{\sin^{n + 2} x} \\[8pt]
+        &= -\frac{\cos x}{\sin^{n + 1} x} -
+            (n + 1) \int \frac{\dif x}{\sin^{n + 2} x} +
+            (n + 1) \int \frac{\dif x}{\sin^n x}
+    \end{align*}
+
+    Пусть $\displaystyle J_n = \int \frac{\dif x}{\sin^n x}$.
+
+    Переобозначим $n = n + 2$ в полученном интеграле, чтобы формула получилась красивой
+    \begin{align*}
+        J_{n - 2} &= -\frac{\cos x}{\sin^{n - 1} x} - (n - 1) J_n + (n - 1) J_{n - 2}\\[8pt]
+        (n - 1)J_n &= -\frac{\cos x}{\sin^{n - 1} x} + (n - 2)J_{n - 2}\\[8pt]
+        J_n &= \frac{\cos x}{(1 - n) \sin^{n - 1} x} + \frac{n - 2}{n - 1}J_{n - 2}
+    \end{align*}
+
+    Тогда:
+    \begin{align*}
+        I_n = \int \frac{\dif x}{(a\sin x + b\cos x)^n}
+        &= \frac{1}{r} \int \frac{\dif x}{(\frac{a}{r}\sin x + \frac{b}{r}\cos x)^n}
+            &\explain{
+                \displaystyle r = \sqrt{a^2 + b^2}
+            } \\[8pt]
+        &= \frac{1}{r} \int \frac{\dif x}{(\cos \phi \sin x + \sin \phi \cos x)^n}
+            &\explain{
+                \displaystyle \phi = \arccos \frac{a}{r} = \arcsin \frac{b}{r}
+            } \\[8pt]
+        &= \frac{1}{r} \int \frac{\dif x}{(\sin (\phi + x))^n} \\[8pt]
+        &= \frac{\cos (\phi + x)}{r(1 - n) \sin^{n - 1} (\phi + x)} +
+            \frac{n - 2}{n - 1} I_{n - 2} \\[8pt]
+        &= \frac{\cos (\arccos{\frac{a}{\sqrt{a^2 + b^2}}} + x)}
+                {\sqrt{a^2 + b^2}(1 - n) \sin^{n - 1} (\arccos{\frac{a}{\sqrt{a^2 + b^2}}} + x)} +
+            \frac{n - 2}{n - 1} I_{n - 2}
+    \end{align*}
+
+\end{document}
diff --git a/calc/sol0203.tex b/calc/sol0203.tex
new file mode 100644
index 0000000..f568637
--- /dev/null
+++ b/calc/sol0203.tex
@@ -0,0 +1,282 @@
+\documentclass[10pt,a5paper]{article}
+\usepackage[svgnames, rgb]{xcolor}
+
+\input{intro}
+
+\lhead{\color{gray} Шарафатдинов Камиль 192}
+\rhead{\color{gray} \texttt{sol0203}}
+\title{ДЗ на 10.02}
+\author{Шарафатдинов Камиль БПМИ-192}
+\date{билд: \today}
+
+
+% -- Here bet  dragons --
+\begin{document}\thispagestyle{empty}
+
+\maketitle
+\clearpage
+\setcounter{page}{1}
+
+
+\newcommand{\deft}{\texttt{\\deft is undefined}}
+    
+
+\question{1.b}{
+    \[
+        \int \frac{\dif x}{\sqrt[6]{(x - 7)^7(x - 5)^5}}
+        = -3\sqrt[6]{\dfrac{x - 5}{x - 7}} + C
+    \]
+}
+
+    \begin{align*}
+        \int \frac{\dif x}{\sqrt[6]{(x - 7)^7(x - 5)^5}}
+            &= -6 \int \frac{t^7 \dif t}{t^7 \br{ \sqrt[6]{x - 5} }^5}
+                &\explain{
+                    \displaystyle t = \sqrt[6]{\frac{1}{x - 7}}\\[8pt]
+                    \displaystyle \frac{\dif t}{\dif x}
+                        = -\frac{1}{6} \br{ \sqrt[6]{\frac{1}{x - 7}} }^7
+                        = -\frac{1}{6} t^7
+                }\\[8pt]
+            &= -6 \int \frac{\dif t}{\br{ \sqrt[6]{x - 5} }^5}\\[8pt]
+            &= -6 \int \frac{t^5 \dif t}{\br{ \sqrt[6]{1 + 2t^6} }^5}
+                &\explain{
+                    \dfrac{1}{\sqrt[6]{x - 5}}
+                        &= \displaystyle \sqrt[6]{\frac{1}{\frac{1}{t^6} + 2}}\\[8pt]
+                        &= \dfrac{t}{\sqrt[6]{1 + 2t^6}}
+                }\\[8pt]
+            &= 3 \int \frac{u^5 \dif u}{u^7}
+             = 3 \int \frac{\dif u}{u^2}
+                &\explain{
+                    \displaystyle u = \dfrac{1}{\sqrt[6]{1 + 2t^6}}\\
+                    \displaystyle \frac{\dif u}{\dif t} = -2t^5 u^7\\
+                }\\
+            &= -\frac{3}{u} + C\\
+            &= -3\sqrt[6]{1 + 2t^6} + C\\
+            &= -3\sqrt[6]{1 + \dfrac{2}{x - 7}} + C\\
+            &= -3\sqrt[6]{\dfrac{x - 5}{x - 7}} + C
+    \end{align*}
+
+\clearpage
+
+\question{7.c}{
+    \[
+        \int \frac{\dif x}{1 + \sqrt{1 - 2x - x^2}}
+    \]
+}
+
+    Воспользуемся почти подстановкой Эйлера:
+    $
+        \displaystyle \sqrt{ax^2 + bx + c} = xt - \sqrt{c}
+    $
+
+    \begin{align*}
+        \int \frac{\dif x}{1 + \sqrt{1 - 2x - x^2}}
+            &= \int \frac{\dif x}{xt}
+                &\explain{
+                    \displaystyle \sqrt{1 - 2x - x^2} = xt - 1\\[4pt]
+                    \displaystyle 1 - 2x - x^2 = x^2t^2 - 2xt + 1\\[4pt]
+                    \displaystyle x \br{ t^2 + 1 } = t - 1\\[4pt]
+                    \displaystyle x = \dfrac{t - 1}{t^2 + 1}\\[16pt]
+                    \displaystyle \dfrac{\dif x}{\dif t} = -2 \dfrac{t^2 - 2t - 1}{(t^2 + 1)^2}
+                }\\
+            &= \int \frac{-2\dfrac{t^2 - 2t - 1}{(t^2 + 1)^2} \dif t}
+                         {2\dfrac{t - 1}{t^2 + 1} t}\\[8pt]
+            &= -\int \frac{(t^2 - 2t - 1)(t^2 + 1) \dif t}
+                          {(t - 1)(t^2 + 1)^2 t}\\[8pt]
+            &= -\int \frac{(t^2 - 2t - 1) \dif t}
+                          {(t - 1)(t^2 + 1) t}\\[8pt]
+            &= -\int \br{
+                \frac{2}{t^2 + 1} + \frac{1}{t} - \frac{1}{t - 1}
+            } \dif t\\[8pt]
+            &= -2\arctan t - \log{t} + \log(t - 1) + C\\[8pt]
+            &= -2\arctan \dfrac{\sqrt{1 - 2x - x^2} + 1}{x} + \log \dfrac{t - 1}{t} + C\\[8pt]
+            &= -2\arctan \dfrac{\sqrt{1 - 2x - x^2} + 1}{x}
+                 + \log \dfrac{\sqrt{1 - 2x - x^2} + 1 - x}{\sqrt{1 - 2x - x^2} + 1} + C\\[8pt]
+    \end{align*}
+
+\clearpage
+
+\question{10.b}{
+    \[
+        \int \frac{x^8 \dif x}{\sqrt{x^2 + 1}}
+    \]
+}
+
+    Лемма (вообще говоря, это задача 9):
+    \begin{gather*}
+        P \in \mathbb{R}_n[x], \qquad
+        Q \in \mathbb{R}_{n - 1}[x], \qquad
+        R = \sqrt{ax^2 + bx + c} \implies
+        \int \frac{P \dif x}{R} = Q R + \lambda \int \frac{\dif x}{R}\\[16pt]
+        \br{ QR + \lambda \int \frac{\dif x}{R}}' =
+        Q'R + QR' + \frac{\lambda}{R} =
+        \frac{Q'R^2}{R} + \frac{Q(2ax + b)}{2R} + \frac{\lambda}{R} =
+        \frac{Q'R^2 + \frac{1}{2} Q(2ax + b) + \lambda}{R}
+    \end{gather*}
+
+    Надо бы ещё доказать, что такое $Q$ всегда найдется, но нам достаточно того, что в задаче такой $Q$ есть.
+
+    Тогда по лемме нам надо разложить $x^8$ на слагаемые $Q'(x^2 + 1) + Qx + \lambda$
+    для некоторого $Q$.
+
+    Пусть $Q = a_7x^7 + \ldots + a_0, \quad Q' = 7a_7x^6 + \ldots + a_1$
+
+    Получится система линейных уравнений, которую я выписывать не буду, а выпишу сразу ответ:
+    \[
+        Q = \frac{1}{8}x^7 - \frac{7}{48} x^5 + \frac{35}{192} x^3 - \frac{35}{128} x, \quad \lambda = \frac{35}{128}
+    \]
+
+    Тогда
+
+    \begin{align*}
+        \int \frac{x^8 \dif x}{\sqrt{x^2 + 1}} &= \br{
+            \frac{1}{8}x^7 - \frac{7}{48} x^5 + \frac{35}{192} x^3 - \frac{35}{128} x
+        } \sqrt{x^2 + 1} + \frac{25}{128}\int \frac{\dif x}{\sqrt{x^2 + 1}} \\[8pt]&=
+        \br{
+            \frac{1}{8}x^7 - \frac{7}{48} x^5 + \frac{35}{192} x^3 - \frac{35}{128} x
+        } \sqrt{x^2 + 1} + \frac{25}{128} \log \left|x + \sqrt{x^2 + 1}\right| + C
+    \end{align*}
+
+\question{17.b}{
+    \[
+        \int \frac{(x + 3) \dif x}{(x^2 + 1)\sqrt{x^2 + x + 1}}
+    \]
+}
+
+    \newcommand{\brac}[2]{ \br{ \dfrac{#1}{#2} } }
+    \newcommand{\sbrac}[2]{ \br{ \frac{#1}{#2} } }
+    \renewcommand{\deft}{\br{ \sqrt{x^2 + x + 1} - x }}
+
+    \begin{gather*}
+        \sqrt{x^2 + x + 1} = x + t\\
+        x^2 + x + 1 = x^2 + 2xt + t^2
+    \end{gather*}
+    \begin{align*}
+        x &= \frac{t^2 - 1}{1 - 2t}\\
+        \dif x &= -\frac{2(t^2 - t + 1)}{(1 - 2t)^2} \dif t\\
+        x + t &= \frac{t^2 - t + 1}{2t - 1}\\
+        x + 3 &= \frac{t^2 - 6t + 2}{1 - 2t}\\
+        x^2 + 1 &= \frac{t^4 + 2t^2 - 4t + 2}{(1 - 2t)^2}
+    \end{align*}
+
+    \begin{align*}
+        \int \frac{(x + 3) \dif x}{(x^2 + 1)\sqrt{x^2 + x + 1}}
+            &= -2 \int \frac{(x + 3)(t^2 - t + 1)\dif t}{(x^2 + 1)(x + t)(1 - 2t)^2}\\[8pt]
+            &= -2 \int \frac{
+                \brac{t^2 - 6t + 2}{1 - 2t}(t^2 - t + 1) \dif t
+            }{
+                \brac{t^4 + 2t^2 - 4t + 2}{(1 - 2t)^2}
+                \brac{t^2 - t + 1}{2t - 1}
+                (1 - 2t)^2
+            }\\[8pt]
+            &= -2 \int - \frac{t^2 - 6t + 2}{t^4 + 2t^2 - 4t + 2} \dif t\\[8pt]
+            &= 2 \int \frac{t^2 - 6t + 2}{t^4 + 2t^2 - 4t + 2} \dif t\\[8pt]
+            &= \frac{1}{\sqrt{2}} \int \br{
+                \frac{2 t - 3 \sqrt2 + 4}{-t^2 + \sqrt2 t + \sqrt2 - 2}
+                + \frac{2 t + 3 \sqrt2 + 4}{t^2 + \sqrt2 t + \sqrt2 + 2}
+            } \dif t\\[8pt]
+    \end{align*}
+
+    \renewcommand{\deft}{t}
+    \def\firstdenum{\br{ \deft - \frac{\sqrt2}{2} }^2 + \sbrac{\sqrt{6 - 4\sqrt2}}{2}^2}
+    \def\firstpoly{ \deft^2 - \sqrt2 \deft - \sqrt2 + 2 }
+    \def\firstsqrt{ \sqrt{6 - 4\sqrt2} }
+
+    \def\seconddenum{ \br{ \deft + \frac{\sqrt2}{2} }^2 + \sbrac{\sqrt{6 + 4\sqrt2}}{2}^2 }
+    \def\secondpoly{ \deft^2 + \sqrt2 \deft + \sqrt2 + 2 }
+    \def\secondsqrt{ \sqrt{6 + 4\sqrt2}  }
+
+    \begin{align*}
+        \int \frac{2 t - 3 \sqrt2 + 4}{-t^2 + \sqrt2 t + \sqrt2 - 2}\dif t
+            &= -2 \int \frac{t - \frac{3\sqrt2}{2} + 2}{\firstdenum}\dif t\\[8pt]
+            &= -2 \int \frac{t - \frac{\sqrt2}{2}}{\firstdenum} \dif t
+               -2 \int \frac{-\sqrt2 + 2}{\firstdenum} \dif t\\[8pt]
+            &= - \int \frac{\dif \br{ \firstpoly }}{\firstpoly} \dif t
+               +2(\sqrt2 - 2) \frac{2}{\firstsqrt} \arctan \frac{2t - \sqrt2}{\firstsqrt}\\[8pt]
+            &= -\log{ \left|\firstpoly\right| } -
+                4\arctan \frac{2t - \sqrt2}{\firstsqrt} + C_1
+    \end{align*}
+
+    \begin{align*}
+        \int \frac{2 t + 3 \sqrt2 + 4}{\secondpoly}\dif t
+            &= \int \frac{2t + 3\sqrt2 + 4}{\seconddenum} \dif t\\[8pt]
+            &= \int \frac{2t + \sqrt2}{\seconddenum} +
+               \int \frac{2\sqrt2 + 4}{\seconddenum}\\[8pt]
+            &= \log{ \left|\secondpoly\right| } +
+               2(\sqrt2 + 2)\frac{2}{\secondsqrt} \arctan \frac{2t + \sqrt2}{\secondsqrt} + C_2\\[8pt]
+            &= \log{ \left|\secondpoly\right| } +
+                4\arctan \frac{2t + \sqrt2}{\secondsqrt} + C_2
+    \end{align*}
+
+    \begin{align*}
+        \int \frac{(x + 3) \dif x}{(x^2 + 1)\sqrt{x^2 + x + 1}}
+            &= -\log{ \left|\firstpoly\right| } -
+                4\arctan \frac{2t - \sqrt2}{\firstsqrt}\\[8pt]
+            &  +\log{ \left|\secondpoly\right| } +
+                4\arctan \frac{2t + \sqrt2}{\secondsqrt} + C\\[8pt]
+    \end{align*}
+    \renewcommand{\deft}{\br{\sqrt{x^2 + x + 1} - x}}
+    \begin{align*}
+            &= -\log{ \left|\firstpoly\right| } -
+                4\arctan \frac{2\deft - \sqrt2}{\firstsqrt}\\[8pt]
+            &  +\log{ \left|\secondpoly\right| } +
+                4\arctan \frac{2\deft + \sqrt2}{\secondsqrt} + C
+    \end{align*}
+
+\clearpage
+\question{17.c}{
+    \[
+        \int \frac{\dif x}{(x^3 - x)\sqrt{x^2 + x + 4}}
+    \]
+}
+    \begin{gather*}
+        \sqrt{x^2 + x + 4} = x + t\\
+        x^2 + x + 4 = x^2 + 2xt + t^2
+    \end{gather*}
+    \begin{align*}
+        x &= \frac{t^2 - 4}{1 - 2t}\\
+        \dif x &= -\frac{2(t^2 - t + 4)}{(1 - 2t)^2} \dif t\\
+        x + t &= \frac{t^2 - t + 4}{2t - 1}
+    \end{align*}
+
+    \medskip
+
+    \renewcommand{\deft}{\sqrt{x^2 + x + 4} - x}
+
+    \begin{align*}
+        \int \frac{\dif x}{(x^3 - x)\sqrt{x^2 + x + 4}}
+            &= \int \frac{\dif x}{x(x + 1)(x - 1)\sqrt{x^2 + x + 4}}\\[8pt]
+            &= -\int \frac{2(t^2 - t + 4) \dif t}{(1 - 2t)^2 x(x + 1)(x - 1)(x + t)}\\[8pt]
+            &= -2\int \frac{(t^2 - t + 4) \dif t}{
+                (1 - 2t)^2
+                \brac{t^2 - 4}{1 - 2t}
+                \brac{t^2 - 5 + 2t}{1 - 2t}
+                \brac{t^2 - 3 - 2t}{1 - 2t}
+                \brac{t^2 - t + 4}{2t - 1}
+            }\\[8pt]
+            &= 2 \int \frac{(1 - 2t)^2 \dif t}{
+                (t^2 - 4)(t^2 + 2t - 5)(t^2 - 2t - 3)
+            }\\[8pt]
+            &= 2 \int \frac{(1 - 2t)^2 \dif t}{
+                (t - 2)(t + 2)(t - \sqrt{6} + 1)(t + \sqrt{6} + 1)(t + 1)(t - 3)
+            }\\[8pt]
+            &= 2\int \br{
+                -\frac{1}{t^2 - 4}
+                -\frac{1}{8(t + 1)}
+                +\frac{1}{8(t - 3)}
+                -\frac{1}{4\sqrt{6}(t + \sqrt{6} + 1)}
+                +\frac{1}{4\sqrt{6}(t - \sqrt{6} + 1)}
+            } \dif t\\[8pt]
+            &= -\arctan \frac{t}{2}
+               -\frac{1}{4}\log |t + 1|
+               +\frac{1}{4}\log |t - 3|
+               -\frac{1}{2\sqrt{6}}\log |t + \sqrt{6} + 1|
+               +\frac{1}{2\sqrt{6}}\log |t - \sqrt{6} + 1|\\[8pt]
+            &= -\arctan \frac{\deft}{2}
+               -\frac{1}{4}\log |\deft + 1|
+               +\frac{1}{4}\log |\deft - 3|\\[8pt]
+               &-\frac{1}{2\sqrt{6}}\log |\deft + \sqrt{6} + 1|
+               +\frac{1}{2\sqrt{6}}\log |\deft - \sqrt{6} + 1|\\[8pt]
+    \end{align*}
+
+\end{document}
diff --git a/calc/sol0210.tex b/calc/sol0210.tex
new file mode 100644
index 0000000..bb4ad94
--- /dev/null
+++ b/calc/sol0210.tex
@@ -0,0 +1,56 @@
+\documentclass[10pt,a5paper]{article}
+\usepackage[svgnames, rgb]{xcolor}
+
+\input{intro}
+
+\lhead{\color{gray} Шарафатдинов Камиль 192}
+\rhead{\color{gray} \texttt{sol0203}}
+\title{ДЗ на 10.02}
+\author{Шарафатдинов Камиль БПМИ-192}
+\date{билд: \today}
+
+
+% -- Here bet  dragons --
+\begin{document}\thispagestyle{empty}
+
+\maketitle
+\clearpage
+\setcounter{page}{1}
+
+\question{1.f}{
+    \[
+        \int_1^e \sin \log x \dif x
+    \]
+}
+    
+    \begingroup
+    \setlength{\jot}{8pt}
+    \begin{align*}
+        I = \int \sin \log \dif x
+            &= \int e^u \sin u \dif u
+                &\explain{
+                    \displaystyle u = \log x\\
+                    \displaystyle \dif u = \frac{\dif x}{x} = \frac{\dif x}{e^u}
+                }\\
+            &= - e^u \cos u + \int e^u \cos u \dif u\\
+            &= - e^u \cos u + e^u \sin u - \int e^u \sin u \dif u
+    \end{align*}
+    \endgroup
+
+    \begin{flalign*}
+        2I = e^u (\sin u - \cos u) + C\\
+        I = \frac{e^u}{2} (\sin u - \cos u) + C =
+            \frac{x}{2} (\sin \log x - \cos \log x) + C
+    \end{flalign*}
+
+    \[
+        \int_1^e \sin \log x \dif x =
+        \frac{e}{2} (\sin 1 - \cos 1) - \frac{1}{2} (0 - 1) =
+        \frac{e}{2} (\sin 1 - \cos 1) + \frac{1}{2}
+    \]
+
+\vspace*{\fill}
+
+    P.S. Ну проверь хотя бы одну задачу, пожаааалуйста
+
+\end{document}