Задача №1427
Условие
Найти интеграл \(\int\frac{x^3dx}{\sqrt{x-1}}\).
Решение
\[
\int\frac{x^3dx}{\sqrt{x-1}}
=\left[\begin{aligned}& u=\sqrt{x-1};\;x=u^2+1.\\& dx=2udu.\end{aligned}\right]
=\int\frac{\left(u^2+1\right)^3\cdot{2u}du}{u}=\\
=2\int\left(u^6+3u^4+3u^2+1\right)du
=\frac{2u^7}{7}+\frac{6u^5}{5}+2u^3+2u+C=\\
=\frac{2u}{35}\cdot\left(5u^6+21u^4+35u^2+35\right)+C
=\frac{2\sqrt{x-1}}{35}\cdot\left(5x^3+6x^2+8x+16\right)+C
\]
Ответ:
\(\frac{2\sqrt{x-1}}{35}\cdot\left(5x^3+6x^2+8x+16\right)+C\)