From fd0f78f4de5ced34aa68b73dd37910b0db8307bd Mon Sep 17 00:00:00 2001
From: syn <isaqtm@gmail.com>
Date: Tue, 28 Jan 2020 00:39:06 +0300
Subject: Init commit

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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}{
+    \[
+        \int \frac{dx}{x^s} = \frac{1}{(1 - s)x^{s - 1}} + C
+    \]
+}
+    \[
+        \braced{ \frac{1}{(1 - s)x^{s - 1}} + C }' =
+        -\frac{0 - (1 - s)(s - 2)x^{s - 2}}{(1 - s)^2 x^{2s - 2}} =
+        \frac{1}{x^s} \qed
+    \]
+
+\question{Лемма 2}{
+    \[
+        \int \frac{dx}{(x^2 + a^2)^2} = \frac{1}{2a^2} \braced{
+            \frac{x}{x^2 + a^2} + \frac{1}{a}\arctan\frac{x}{a} 
+        } + C
+    \]
+}
+
+    \begin{align*}
+        \braced{
+            \frac{1}{2a^2} \braced{
+                \frac{x}{x^2 + a^2} + \frac{1}{a}\arctan\frac{x}{a}
+            }
+        }' =
+        \frac{1}{2a^2} \braced{
+            \frac{x^2 + a^2 - 2x^2}{(x^2 + a^2)^2} + \frac{1}{x^2 + a^2}
+        } =
+        \frac{1}{2a^2} \braced{
+            \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}\braced{
+            \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}{
+    \[
+        \int \frac{x^5 - x}{x^8 + 1}dx = \frac{\sqrt{2}}{8} \braced{
+            \log |x^4 - \sqrt{2} x^2 + 1| - \log |x^4 + \sqrt{2} x^2 + 1|
+        } + \bar{C}
+    \]
+}
+    
+    \[
+        \int \frac{x^5 - x}{x^8 + 1}dx =
+        \int \frac{x(x^4 - 1}{x^8 + 1}dx =
+        \begin{bmatrix}
+        u = x^2\\
+        dx = \frac{du}{2x}    
+        \end{bmatrix} =
+        \int \frac{x(u^2 - 1)}{u^4 + 1}\frac{du}{2x} =
+        \frac{1}{2} \int \frac{u^2 - 1}{u^4 + 1} du
+    \]
+    \[
+        u^4 + 1 = \braced{u^2 + \sqrt{2} u + 1}\braced{u^2 - \sqrt{2} u + 1}
+    \]
+    \[
+        \frac{u^2 - 1}{u^4 + 1} =
+        \frac{Au + B}{u^2 + \sqrt{2} u + 1} +
+        \frac{Cu + D}{u^2 - \sqrt{2} u + 1}
+    \]
+
+    \[
+        (Au + B)(u^2 - \sqrt{2}u + 1) + (Cu + D)(u^2 + \sqrt{2}u + 1) \equiv u^2 - 1
+    \]
+
+    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 \braced{ 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 \braced{ 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}
+
+    \[
+        \frac{1}{2} \int \frac{u^2 - 1}{u^4 + 1} du =
+        \frac{\sqrt{2}}{8} \braced{
+            \log |u^2 - \sqrt{2} u + 1| - \log |u^2 + \sqrt{2} u + 1|
+        } + C_3
+    \]
+
+    Обратно к $x$:
+    \[
+        \int \frac{x^5 - x}{x^8 + 1}dx =
+        \frac{\sqrt{2}}{8} \braced{
+            \log |x^4 - \sqrt{2} x^2 + 1| - \log |x^4 + \sqrt{2} x^2 + 1|
+        } + \bar{C}
+    \]
+
+\clearpage
+\question{(seminar0113) 13}{
+    \[
+        \int \frac{x \ dx}{(x^2 + 1)(x + 2)(x + 3)} = \frac{1}{20} \braced{
+            -8 \log |x + 2| + 6 \log |x + 3| + \log (x^2 + 1) + 2\arctan x
+        } + \bar{C}
+    \]
+}
+    \[
+        \frac{x}{(x^2 + 1)(x + 2)(x + 3)} =
+        \frac{Ax + B}{x^2 + 1} + \frac{C}{x + 2} + \frac{D}{x + 3}
+    \]
+
+    \[
+        (Ax + B)(x + 2)(x + 3) + C(x^2 + 1)(x + 3) + D(x^2 + 1)(x + 2) \equiv x
+    \]
+    \[\begin{cases*}
+        A = 0.1\\
+        B = 0.1\\
+        C = -0.4\\
+        D = 0.3\\
+    \end{cases*}\]
+
+    \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} \braced{
+                \frac{1}{2}\int \frac{2x \ dx}{x^2 + 1} + \int \frac{dx}{x^2 + 1}
+            } =
+            \frac{1}{10} \braced{
+                \frac{1}{2} \log (x^2 + 1) + \arctan(x)
+            } + C_3
+    \end{gather*}
+
+    \[
+        \int \frac{x \ dx}{(x^2 + 1)(x + 2)(x + 3)} = \frac{1}{20} \braced{
+            -8 \log |x + 2| + 6 \log |x + 3| + \log (x^2 + 1) + 2\arctan x
+        } + \bar{C}
+    \]
+
+\question{(seminar0120) 2.4}{
+    \[
+        \int \frac{dx}{x(x^2 + 1)^2} = -\frac{1}{2} \braced{
+            -\log (x^2 + 1) +
+            \frac{1}{x^2 + 1} +
+            2\log |x|
+        } + \bar{C}
+    \]
+}
+
+    \[
+        \frac{1}{x(x^2 + 1)^2} =
+        \frac{Ax + B}{x^2 + 1} + \frac{Cx + D}{(x^2 + 1)^2} + \frac{E}{x}
+    \]
+
+    \[
+        (Ax + B)(x^2 + 1)x + (Cx + D)x + E(x^2 + 1)^2 \equiv 1
+    \]
+
+    \[\begin{cases*}
+        A = -1\\
+        B = 0\\
+        C = -1\\
+        D = 0\\
+        E = 1
+    \end{cases*}\]
+
+    \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} \braced{
+                \int \frac{(2x - 1) dx}{x^2 - x + 1} -
+                \int \frac{2 dx}{\braced{ x - \frac{1}{2} }^2 + \frac{3}{4}}
+            }\nonumber \\[8pt]
+            &=
+            -\frac{1}{9} \braced{
+                \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} \braced{
+                \int \frac{(2x - 1)dx}{(x^2 - x + 1)^2} -
+                \int \frac{dx}{\braced{ \braced{ x - \frac{1}{2} }^2 + \frac{3}{4} }^2}
+            }\nonumber \\[8pt]
+            &=
+            -\frac{1}{6} \braced{
+                -\frac{1}{x^2 - x + 1} +
+                \frac{2}{3} \braced{
+                    \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} \braced{
+                \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} \braced{
+            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
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