Evaluate Int [(x^4 -1) ^2*4x^3 dx].

Int (x^4-1)^2 * 4x^3 dx

Let u
= x^4  ==> du = 4x^3 dx

==> intg (x^4 -1)^2 (
4x^3) dx = intg (u-1)^2 * du.

Let us integrate with respect
to u.

==> intg (u-1)^2 du = intg (u^2 - 2u + 1)
du

                               = intg u^2 du - intg 2u
du + intg 1 du.

                                = u^3/3 -
2u^2/2  + u + C

==> intg (u-1)^2 du = (1/3)u^3 - u^2
+ u + c

Now we will substitute with u=
x^4

==> intg (u^2-1)du = (1/3)(x^4)^3 - (x^4)^2 +
x^4 + C

                              = (1/3)x^12 - x^8 +
x^4 + C

==> intg (x^4-1)^2 * 3x^2 dx =
(1/3)x^12 - x^8 + x^4 + C