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 ! 
 A: 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. 
A: $$\int{ x^2 \cdot \tan^{-1} x} dx = $$
$$ \tan^{-1} x dx = du $$
$$ x\cdot\tan^{-1} x - \frac{1}{2}\log(x^2+1) = u $$
$$ x^2 = v $$
$$ 2xdx = dv $$
$$\int{ x^2 \cdot \tan^{-1} x} dx = x^2\left(x\cdot\tan^{-1} x - \frac{1}{2}\log(x^2+1) \right)-\int 2x\left(x\cdot\tan^{-1} x - \frac{1}{2}\log(x^2+1) \right)dx$$
$$I = {x^2}\left( {x\cdot{{\tan }^{ - 1}}x - \frac{1}{2}\log ({x^2} + 1)} \right) - 2I + \int {x\log \left( {{x^2} + 1} \right)} dx$$
$$3I = {x^2}\left( {x\cdot{{\tan }^{ - 1}}x - \frac{1}{2}\log ({x^2} + 1)} \right) + \frac{1}{2}\int {\log u} du$$
$$3I = {x^2}\left( {x\cdot{{\tan }^{ - 1}}x - \frac{1}{2}\log ({x^2} + 1)} \right) + \frac{{{x^2} + 1}}{2}\left[ {\log \left( {{x^2} + 1} \right) - 1} \right]$$
$$I = \frac{{{x^2}}}{3}\left( {x\cdot{{\tan }^{ - 1}}x - \frac{1}{2}\log ({x^2} + 1)} \right) + \frac{{{x^2} + 1}}{6}\left[ {\log \left( {{x^2} + 1} \right) - 1} \right]$$
This simplifies to $$I = \frac{{{x^3}}}{3}\cdot{\tan ^{ - 1}}x + \frac{1}{6}\log \left( {{x^2} + 1} \right) - \frac{{{x^2}}}{6} + C $$
which coincides with WA's solution. 
A: Suggestion: Since $\frac{\mathrm{d}}{\mathrm{d}x}\arctan(x)=\frac{1}{1+x^2}$, it seems to be simpler to let $u=\arctan(x)$ and $\mathrm{d}v=x^2\,\mathrm{d}x$. Then you get
$$
\begin{align}
\frac{x^3}{3}\arctan(x)-\frac13\int\frac{x^3}{1+x^2}\mathrm{d}x
&=\frac{x^3}{3}\arctan(x)-\frac13\int\left(x-\frac{x}{1+x^2}\right)\mathrm{d}x\\
&=\frac{x^3}{3}\arctan(x)-\frac16x^2+\frac16\int\frac{\mathrm{d}(1+x^2)}{1+x^2}\\
&=\frac{x^3}{3}\arctan(x)-\frac16x^2+\frac16\log(1+x^2)+C
\end{align}
$$
A: 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
$$
\begin{align}
\int u\;dv & = uv - \int v\;du = \frac{x^3}{3}\arctan x - \int \frac{x^3\;dx}{3(x^2+1)} = \frac{x^3}{3}\arctan x - \frac 1 3\int \left(x- \frac{x}{x^2+1}\right)\;dx \\  \\
& = \frac{x^3}{3}\arctan x - \frac{x^2}{6} + \frac 16 \log(x^2+1) +C.
\end{align}
$$
