Задача №1405
Условие
Найти интеграл \(\int\frac{x^3dx}{\sqrt{1+x^2}}\).
Решение
\[
\int\frac{x^3dx}{\sqrt{1+x^2}}
=\left[\begin{aligned}& u=x^2;\,du=2xdx.\\& dv=\frac{xdx}{\sqrt{1+x^2}}=\frac{1}{2}\left(1+x^2\right)^{-\frac{1}{2}}d\left(1+x^2\right);\,v=\sqrt{x^2+1}.\end{aligned}\right]=\\
=x^2\sqrt{x^2+1}-\int\sqrt{x^2+1}\cdot{2x}dx
=x^2\sqrt{x^2+1}-\int\left(x^2+1\right)^{\frac{1}{2}}d\left(x^2+1\right)
=x^2\sqrt{x^2+1}-\frac{2\sqrt{\left(x^2+1\right)^3}}{3}+C
\]
Ответ:
\(x^2\sqrt{x^2+1}-\frac{2\sqrt{\left(x^2+1\right)^3}}{3}+C\)