\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}