Question

Integral of 13y/(y^4+1).
Can someone explain the process of solving this?

Answer 1

sub \(u =y^2\)

Answer 2

use trigo sub :
y^2 = tan(theta)

Answer 3

y^2 = tanθ
2y dy = sec^2 (θ) dθ

so,
int 13y/(y^4 +1) dy
= 13/2 int (sec^2 θ)/(tan^2 θ + 1) dθ
= 13/2 int (sec^2 θ)/(sec^2 θ) dθ
= 13/2 int dθ
= ....

Answer 4

ahhhh okay, thanks! :)

Answer 5

welcome :)