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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