∫▒x^3/(1+x^2 ) dx

This is the question I have. It's quite a mind boggler (well, for me)

It has to do with further substitution with integration.

This is what I did, but I got the wrong answer in the back of the book.

Let u = 1+x^2

du/dx = 2x

du/2x = dx

∫〖(x^3/u)(du/2x)〗

∫((u-1)/2u du)

∫〖1/2-1/2u〗

=1/2 u- 1/2 lnu

=1/2(1+x^2 )-1/2 ln(1+x^2 )+c

I have a feeling it's something with my substitution