Integration of arctan(x) is itself?

You can't take the $x$ out of the first integral on the right side because $x$ is the integration variable and is not a constant. You would have to handle it differently like this: Let $u = 1+x^2$, and can you continue to the finish?

Just so this question can be answered:

You originally did the integration by parts incorrectly:

$u=x,\ \ \ \ \ dv=\frac{1}{1+x^2}dx$

$du=dx,\ \ \ \ \ v=\int\frac{1}{1+x^2}dx$

$$\int\frac{x}{1+x^2}dx=x\int\frac{1}{1+x^2}dx-\int\left(\int\frac{1}{1+x^2}dx \right)dx$$

we have $$\int \frac{x}{1+x^2}dx=\frac{1}{2}\int \frac{2x}{1+x^2}dx$$