Question

Find the integral of `(x^2+1)/(x^3+3x)` ?

Answer 1

`I=1/3ln(abs(x^3+3x))+C`

Explanation:

`I= int(x^2+1)/(x^3+3x) dx`

Let `u=x^3+3x -> dx=1/(3x^2+3) du = 1/3 * 1/(x^2+1) du`

`:. I=1/3 int 1/u du`

`= 1/3 ln abs(u) +C`

Undo substitution

`I= 1/3ln(abs(x^3+3x))+C`