Indefinite integral of $\frac{\arctan x}{x^2+1}$

Question

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!

Answer 1

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

Answer 2

Hint

Notice that

$$\int f'f=\frac12 f^2+c$$ so the answer for your antiderivative is $\frac12(\arctan x)^2+c$.

Answer 3

Your notation may be a bit off?

Doing the $u$-substitution gives

$$\int \frac{\arctan x}{1+x^2}dx = \int u\text{ }du = \frac{u^2}{2} + C = \frac{(\arctan x)^2}{2} + C.$$

Note that $(\arctan x)^2 \neq \arctan \left(x^2\right)$.