Evaluating the indefinite integral $\int\frac{x\arctan x}{(1+x^2)^2}dx$ The integral again is $$\int\frac{x\arctan x}{(1+x^2)^2}dx$$
I'm actually pretty much just stumped on this one. It looks like it should be amenable to a substitution kinda like $\arctan x$
but that obviously does not work. 
I also thought maybe by parts, since 
$$
\int\frac{\arctan x}{1+x^2}dx
$$
Isn't too bad to anti differentiate but this approach seems to complicate things on the differentiation side. 

Here is a way by direct integration by parts:
Let $u = \arctan x, dv = \frac{x}{(1+x^2)^2}dx$, then $du = \frac{1}{x^2+1}dx, v = -\frac{1}{2(x^2+1)}$.
Then  $$\int\frac{x\arctan x}{(1+x^2)^2}dx =-\frac{1}{2(x^2+1)}\arctan x+\int   \frac{1}{2(x^2+1)}\frac{1}{x^2+1}dx=\int   \frac{1}{2(x^2+1)^2}dx$$
Try some trigonometric subsititution for $\int   \frac{1}{2(x^2+1)^2}dx$
