How to find$\int x^2 \arctan(x) \;dx $ 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 ! 

We have
$$
\int x^2 \arctan x\;dx.
$$
Let
$$
\begin{align}
u & = \arctan x, \\  \\
du & = \frac{dx}{x^2+1}, \\  \\
dv & = x^2 \;dx, \\  \\
v & = \frac{x^3}{3}.
\end{align}
$$
Then
$$
\begin{align}
\int u\;dv & = uv - \int v\;du = \frac{x^3}{3}\arctan x - \int \frac{x^3\;dx}{3(x^2+1)} = \frac{x^3}{3}\arctan x - \frac 1 3\int \left(x- \frac{x}{x^2+1}\right)\;dx \\  \\
& = \frac{x^3}{3}\arctan x - \frac{x^2}{6} + \frac 16 \log(x^2+1) +C.
\end{align}
$$
