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 ! 

Let $u=\arctan x$, $dv = x^2\,dx$. Then $du = \frac{dx}{1+x^2}$, $v = \frac{1}{3}x^3$, so
$$\begin{align*}
\int x^2\arctan x\,dx &= \frac{1}{3}x^3\arctan x - \frac{1}{3}\int\frac{x^3}{1+x^2}\,dx\\
&= \frac{1}{3}x^3\arctan x - \frac{1}{3}\int\left(\frac{x^3+x}{1+x^2}-\frac{x}{1+x^2}\,dx\right)\\
&= \frac{1}{3}x^3\arctan x - \frac{1}{3}\int x\,dx + \frac{1}{3}\int\frac{x}{1+x^2}\,dx\\
&= \frac{1}{3}x^3\arctan x - \frac{1}{6}x^2 +\frac{1}{3}\int\frac{\frac{1}{2}\,du}{u} &\quad&(u=1+x^2)\\
&=\frac{1}{3}x^3\arctan x - \frac{1}{6}x^2 + \frac{1}{6}\int\frac{du}{u}\\
&= \frac{1}{3}x^3\arctan x - \frac{1}{6}x^2 + \frac{1}{6}\ln|u| + C\\
&= \frac{1}{3}x^3\arctan x - \frac{1}{6}x^2 + \frac{1}{6}\ln|1+x^2|+C\\
&= \frac{1}{3}x^3\arctan x - \frac{1}{6}x^2 + \frac{1}{6}\ln(1+x^2)+C.
\end{align*}$$
If after integration by parts/substitution, the resulting integral is harder than the one you started with, then it's time to go back and try a different integration by parts/substitution. 
