I was solving $\int x^2 \arctan(x) \;dx $ I set $u=x^2$, $dv= \arctan(x)$, so I could get $du=2x$, $v=x\arctan(x)-\frac12\ln(1+x^{2})$.
From $\int x^2 \arctan(x)\;dx = uv - \int v \; du$
I got
$\begin{align*}&\int x^2 \arctan(x) \; dx =\\ &x^3\arctan(x) -\frac12\ln(1+x^{2})-\int 2x\left[x^{2}\arctan x -\frac12\ln(1+x^2)\right]\;dx \end{align*}$
and simplified if; then I got
$3\int x^2\arctan(x) \;dx = x^3 \arctan(x)-\frac12\ln(1+x^2)+\int x\ln(1+x^2) \;dx$
after that I use $w=\ln(1+x^2) \; dv =dx$ to find $\int x\ln(1+ x^2)$
but I got
$x^2 \arctan(x)-\int 2x\arctan(x) \; dx$
If I got $x^2 \arctan(x)-\int 2x^2 \arctan(x)\;dx$ instead, it would be easy to solve the question....
How can I solve this question and if you find any my mistake could you post this wall ??
Thank you !