Задача №1597
Условие
Найти интеграл \(\int\frac{\left(x^4+1\right)dx}{x^3-x^2+x-1}\).
Решение
\[
\frac{x^4+1}{x^3-x^2+x-1}
=x+1+\frac{2}{x^3-x^2+x-1}
\]
\[
\frac{2}{(x-1)\left(x^2+1\right)}
=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}
=\frac{A\left(x^2+1\right)+(Bx+C)(x-1)}{(x-1)\left(x^2+1\right)}
\]
\[
2=A\left(x^2+1\right)+(Bx+C)(x-1)
\]
\[
\begin{aligned}
& x=1;\;2=2A;\;A=1.\\
& x=0;\;2=A-C=1-C;\;C=-1.\\
& x=2;\;2=5A+2B+C=4+2B;\;B=-1.
\end{aligned}
\]
\[
\int\frac{\left(x^4+1\right)dx}{x^3-x^2+x-1}
=\int\left(x+1+\frac{1}{x-1}-\frac{x}{x^2+1}-\frac{1}{x^2+1}\right)dx=\\
=\frac{x^2}{2}+x+\ln|x-1|-\frac{1}{2}\ln\left(x^2+1\right)-\arctg{x}+C
\]
Ответ:
\(\frac{x^2}{2}+x+\ln|x-1|-\frac{1}{2}\ln\left(x^2+1\right)-\arctg{x}+C\)