Задача №1582
Условие
Найти интеграл \(\int\frac{x^3+1}{x^3-x^2}dx\).
Решение
\[
\frac{x^3+1}{x^3-x^2}
=\frac{x^3-x^2+x^2+1}{x^3-x^2}
=1+\frac{x^2+1}{x^3-x^2}
\]
\[
\frac{x^2+1}{x^2(x-1)}
=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}
=\frac{Ax(x-1)+B(x-1)+Cx^2}{x^2(x-1)}
\]
\[
x^2+1
=Ax(x-1)+B(x-1)+Cx^2
\]
\[
\begin{aligned}
& x=0;\;B=-1.\\
& x=1;\;C=2.\\
& x=2;\;5=2A+B+4C;\;A=-1.
\end{aligned}
\]
\[
\int\frac{x^3+1}{x^3-x^2}dx
=\int\left(1-\frac{1}{x}-\frac{1}{x^2}+\frac{2}{x-1}\right)dx
=x-\ln|x|+\frac{1}{x}+2\ln|x-1|+C
\]
Ответ:
\(x+\frac{1}{x}-\ln|x|+2\ln|x-1|+C\)