Integral of $\arctan x$ 

This could be integrated using integration by parts. Let 
\[u=\arctan x,   dv=dx   \Rightarrow  du=\frac{1}{1+x^2},  v=x.\]
Using integration by parsts
\[\int u dv= uv -\int vdu.\]
Hence
\[(1)    \int \arctan x  dx = x \arctan x - \int \frac{x}{1+x^2}dx.\]
Substitute $u=1+x^2$. Then $du=2x dx$ and hence
\[\int \frac{x}{1+x^2}dx=\int \frac{\frac{1}{2} du}{u}=\frac{1}{2}\ln |u|+c=\frac{1}{2}\ln (1+x^2). \]
Hence from (1) we get 
\[\int \arctan x dx = x \arctan x - \frac{1}{2}\ln (1+x^2)+c.\]