I tried to solve the integral $\int\frac{\arctan{x}}{1+x^2}dx$ and I got lost. I have the answer, but I have no idea how to get there.

Question

I do not know how to integrate

$$\int{\arctan{x} \over 1+x^2}dx$$

The answer is given to be: $\arctan{x} + 1/2 \ln{|1+x^2|}+C$

I'd appreciate your help very much. Thanks!

Edit 1: it must be done with substitution only.

Edit 2: Yes. It's not the answer. Thank you all. I'll delete this in a minute. Thanks!!

Answer 1

If you set $u=\arctan{x}$ you get $du={dx\over 1+x^2}$ and the integral rewrites as

$$\int udu={u^2\over 2}+C$$

And I share the trouble of Andre Nicolas and Claude Leibovici and it has nothing to do with the fact that the three of us are French

If you derive the "answer" that's given you get

$${1+x\over 1+x^2}\neq {\arctan{x}\over 1+x^2}$$

Answer 2

Notice that $ff'$ is the derivative of $\frac12 f^2+C$ and in your example $f(x)=\arctan x$.