Задача №1632
Условие
Найти интеграл \(\int\frac{x^3}{(x-1)^{12}}dx\).
Решение
\[
\int\frac{x^3}{(x-1)^{12}}dx
=[u=x-1]
=\int\frac{(u+1)^3du}{u^{12}}=\\
=\int\left(u^{-9}+3u^{-10}+3u^{-11}+u^{-12}\right)du
=-\frac{1}{8(x-1)^8}-\frac{1}{3(x-1)^9}-\frac{3}{10(x-1)^{10}}-\frac{1}{11(x-1)^{11}}+C
\]
Ответ:
\(-\frac{1}{8(x-1)^8}-\frac{1}{3(x-1)^9}-\frac{3}{10(x-1)^{10}}-\frac{1}{11(x-1)^{11}}+C\)