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diff --git a/calc/sol0120.tex b/calc/sol0120.tex new file mode 100644 index 0000000..f073973 --- /dev/null +++ b/calc/sol0120.tex @@ -0,0 +1,372 @@ +\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}
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