What is the integral of (x/(1+x^2))dx?

Answer 1

Let u = 1+x^2
du = 2x dx

Int ((x/(1+x^2))dx = 1/2 int(du/u) = 1/2 ln u = 1/2ln(1+x^2)

Answer 2

Let u = 1+x^2
Then du = 2xdx
So int xdx/(1+x^)dx =1/2 int du/u = 1/2 ln(1+x^2) +C