Интеграл (x^2) *arctg (x)dx

$\displaystyle I=\int x^2\arctan x\, dx\\ \\ I=\dfrac{1}{3}\int \arctan x\, d(x^3)\\ \\ I=\dfrac{1}{3}\left(x^3\arctan x-\int x^3\cdot \dfrac{1}{1+x^2}\,dx\right)\\ \\ I=\dfrac{1}{3}\left(x^3\arctan x-\int x-\dfrac{x}{1+x^2}\,dx\right)\\ \\ I=\dfrac{1}{3}\left(x^3\arctan x-\dfrac{x^2}{2}-\dfrac{1}{2}\ln(1+x^2)\right)\\ \\ \\ \boxed{I=\dfrac{x^3\arctan x}{3}-\dfrac{x^2}{6}-\dfrac{1}{6}\ln(1+x^2)+C}$