How to evaluate these 2 indefinite integrals?

Question

I will strongly appreciate your help:
1) integral ln^3/4dx (ln in power of 3over4)
2)integral 6x^2 / 1-x^2dx (6x^2 over 1-x^2)

The first integral is lnx^3/4, not ln^3/4.My mistake.

Answer 1

1) ln^3/4 doesn't make sense

2)
... 6 x^2 / (1 - x^2) =
... ... = 6 / (1 - x^2) - 6
... ... = 6 / (1 - x)(1 + x) - 6
... ... = 3 / (1 - x) + 3 / (1 + x) - 6

The antiderivative is
... 3 ln |x - 1| + 3 ln |x + 1| - 6 x + C

Answer 2

The integral of Ln(x^(3/4))dx is (3/4)(x*Ln(x)-x) + C.
The integral of Ln(x)^(3/4)dx is
(1/4)*(4x+(3*Gamma[3/4,-Ln(x)])/(-Ln(x))^(3/4))*Ln(x) + C
where Gamma is the incomplete Euler gamma function.

I don't know which of these you meant to write.