Задача №1590
Условие
Найти интеграл \(\int\frac{\left(7x^3-9\right)dx}{x^4-5x^3+6x^2}\).
Решение
\[
\frac{7x^3-9}{x^4-5x^3+6x^2}
=\frac{7x^3-9}{x^2(x-2)(x-3)}=\\
=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-2}+\frac{D}{x-3}
=\frac{Ax(x-2)(x-3)+B(x-2)(x-3)+Cx^2(x-3)+Dx^2(x-2)}{x^2(x-2)(x-3)}.
\]
\[
7x^3-9
=Ax(x-2)(x-3)+B(x-2)(x-3)+Cx^2(x-3)+Dx^2(x-2)
\]
\[
\begin{aligned}
& x=0;\;-9=6B;\;B=-\frac{3}{2}.\\
& x=2;\;47=-4C;\;C=-\frac{47}{4}.\\
& x=3;\;180=9D;\;D=20.\\
& x=1;\;-2=2A+2B-2C-D=2A+\frac{1}{2};\;A=-\frac{5}{4}.
\end{aligned}
\]
\[
\int\frac{\left(7x^3-9\right)dx}{x^4-5x^3+6x^2}
=\int\left(-\frac{5}{4}\cdot\frac{1}{x}-\frac{3}{2}\cdot\frac{1}{x^2}-\frac{47}{4}\cdot\frac{1}{x-2}+\frac{20}{x-3}\right)dx=\\
=-\frac{5}{4}\ln|x|+\frac{3}{2x}-\frac{47}{4}\ln|x-2|+20\ln|x-3|+C
\]
Ответ:
\(-\frac{5}{4}\ln|x|+\frac{3}{2x}-\frac{47}{4}\ln|x-2|+20\ln|x-3|+C\)