Indefinite integral of $\frac{\arctan x}{x^2+1}$ EDIT: I was studying from a site that uses really ambiguous notation so I misread $\arctan\ (x)^2$ as $\arctan\ (x^2)$. Now I can see why the integral is actually $\frac{1}{2} \arctan^2\ x + c $. Thanks to everyone who answered and corrected me!
Why is $$\int\frac{\arctan\ x}{x^2+1} dx=\frac{1}{2}\arctan(x^2)+c$$
instead of $\int u\ du$ where $u=\arctan\ x$ and $du=\frac{1}{x^2+1} dx$, thus $$\begin{align} &\int\frac{\arctan\ x}{x^2+1} dx=\int \arctan\ x\ dx \\
&x\ \arctan\ x-\int \frac{x}{x^2+1}\ dx \\
&x\ \arctan\ x -\frac{1}{2}\ln(x^2+1)+c\end{align}$$
Is there something I'm not getting? I'd like to see the steps for the correct calculation so I can better understand it. Thanks!

instead of $\int u\ du$ where $u=\arctan\ x$ and $du=\frac{1}{x^2+1} dx$, thus $$\int\frac{\arctan\ x}{x^2+1} dx=\int u du= \frac12u^2+C=\frac12(\arctan x)^2+C$$
