Antiderivative of x+1/x^2+1?

Question

I am working on a problem that requires me to integrate x+1/x^2+1. I've used a CAS to find a solution: 1/2ln(x^2+1)+arctan(x), but I would like to know how to get to that answer.

Could anyone explain to me how to find an antiderivative for this problem?

Answer 1

I = ∫ (x + 1) / (x ² + 1) dx
I = (1/2) ∫ (2x / x ² + 1) dx + ∫ ( 1 / (x ² + 1) ) dx
I = (1/2) log (x ² + 1) + tan^(-1) x + C

Answer 2

We have to find
∫ (x+1) dx / (x²+1).
Break the integral in two:
We get
∫ x dx/x+1 = ½∫ 2x dx/(x²+1)
+
∫ dx / (x²+1).
The first integral is of the form ∫ du/u with u = x² +1
and an antiderivative of the second is arctan x.
So the final answer
is ½ ln(x²+1) + arctan x + C.