Question About A Separable Equations Problem

y^3 dy/dx = (y^4 + 1) cos x

so I separate the variables:

y^3 / (y^4 +1) dy = cos x

integrate both sides:

1/4 ln(y^4 +1) = sin x

Can anyone explain to me why the integration of y^3 / (y^4 +1) = 1/4 ln(y^4 +1)
I can see that the integral of 1/(y^4 + 1) = ln(y^4 + 1) but I don't get the 1/4 part.

I kno it has something to do with the substitution rule...but i don't remember that very well

thanks!

Let u = y^4 + 1 => du = 4y^3 dy ...

Ahh I see now.
yeah i need to review substitution...cause that was really simple in retrospect...

Thanks!