What's the indefinite integral of x^3 / (x^4 - 1)

Steps please :)

Using mental substitution,
integral of x^3 / (x^4 - 1) dx
= integral of (1/4) (x^4 - 1)' / (x^4 - 1) dx
= (1/4)ln|x^4 - 1| + c

Consider the derivative of x^4 - 1 which is 4x^3
Multiply and divide our expression by 4
(1/4) ∫4x^3 dx / (x^4 - 1)
Assume x^4 - 1 = t
4x^3 dx = dt
Our integral becomes,
(1/4) ∫ dt / t = (1/4) ln t + c
= (1/4) ln (x^4 - 1) + c

1/4*ln(x^4 - 1)+ C

Hint: Let u = x^4 - 1