Evaluate $\int \arctan({\frac{x-2}{x+2}})\,\text dx$ $$
\int \arctan\left({\frac{x-2}{x+2}}\right)\,\text dx
$$
It seems to be solved in just one line:
$$
x \arctan\left({\frac{x-2}{x+2}}\right)-\log|x^2+4|+c
$$
but how?
I have tried with integration by parts (by picturing out $1$ multiplying arctan), but I don't think is the right way.

We have using first integration by parts:
$$
\begin{aligned}
\int \arctan\frac{x-2}{x+2}\; dx
&=
\int x'\cdot\arctan\frac{x-2}{x+2}\; dx
\\
&=
x\arctan\frac{x-2}{x+2}
-
\int x\cdot\underbrace{\left(\arctan\frac{x-2}{x+2}\right)'}_{2/(x^2+4)}\; dx
\\
&=
x\arctan\frac{x-2}{x+2}
-
\int \frac{d(x^2+4)}{x^2+4}\\
&=
x\arctan\frac{x-2}{x+2}
-
\log(x^2+4)+ c\ .
\end{aligned}
$$
