Question

How do you integrate `x/(x^2+1)`?

Answer 1

`int(x/(x^2+1))dx=1/2ln(x^2+1)+C`

Explanation:

`int(x/(x^2+1))dx`

now `d/(dx)(x^2+1)=2x`

so using `int(f'(x))/(f(x))=ln|f(x)|`

we have `int(x/(x^2+1))dx=1/2ln(x^2+1)+C`