Задача №1588
Условие
Найти интеграл \(\int\frac{x^5dx}{(x-1)^2\left(x^2-1\right)}\).
Решение
\[(x-1)^2\left(x^2-1\right)=x^4-2x^3+2x-1\]
Разделив \(x^5\) на \(x^4-2x^3+2x-1\), получим:
\[
x^5
=\left(x^4-2x^3+2x-1\right)(x+2)+4x^3-2x^2-3x+2
\]
\[
\frac{x^5}{(x-1)^2(x^2-1)}
=x+2+\frac{4x^3-2x^2-3x+2}{(x-1)^2\left(x^2-1\right)}
=x+2+\frac{4x^3-2x^2-3x+2}{(x-1)^3(x+1)}
\]
\[
\frac{4x^3-2x^2-3x+2}{(x-1)^3(x+1)}
=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{(x-1)^3}+\frac{D}{x+1}=\\
=\frac{A(x-1)^2(x+1)+B(x-1)(x+1)+C(x+1)+D(x-1)^3}{(x-1)^3(x+1)}
\]
\[
4x^3-2x^2-3x+2
=A(x-1)^2(x+1)+B(x-1)(x+1)+C(x+1)+D(x-1)^3
\]
\[
\begin{aligned}
& x=1;\;C=\frac{1}{2}.\\
& x=-1;\;D=\frac{1}{8}.\\
& x=0;\;2=A-B+\frac{3}{8}.\\
& x=2;\;20=3A+3B+\frac{13}{8}.
\end{aligned}
\]
Решая систему уравнений \(\left\{\begin{aligned}&A-B=\frac{13}{8};\\&3A+3B=\frac{147}{8}\end{aligned}\right.\), получим \(A=\frac{31}{8}\), \(B=\frac{9}{4}\).
\[
\int\frac{x^5dx}{(x-1)^2\left(x^2-1\right)}
=\int\left(x+2+\frac{31}{8}\cdot\frac{1}{x-1}+\frac{9}{4}\cdot\frac{1}{(x-1)^2}+\frac{1}{2}\cdot\frac{1}{(x-1)^3}+\frac{1}{8}\cdot\frac{1}{x+1}\right)dx=\\
=\frac{x^2}{2}+2x+\frac{31}{8}\ln|x-1|-\frac{9}{4(x-1)}-\frac{1}{(x-1)^2}+\frac{1}{8}\ln|x+1|+C
\]
Ответ:
\(\frac{x^2}{2}+2x+\frac{31}{8}\ln|x-1|-\frac{9}{4(x-1)}-\frac{1}{(x-1)^2}+\frac{1}{8}\ln|x+1|+C\)