∫ (4x+4)(x^2-2x-3)^2/3 dx

Answer 1

I suspect a typo. You probably meant
∫(4x-4)(x^2-2x-3)^(2/3) dx
or
∫(4x+4)(x^2+2x-3)^(2/3) dx

Assuming the first,

Let
u = x^2-2x-3
du = 2x-2 dx

Then you have

∫ 2u^(2/3) du

which is easy.
...
right?

If there is no typo, you are in deep trouble.

Answer 2

Hi Steve, thanks for your reply. Yes, it was a typo and your 2nd assumption was right. Please help, thanks!

∫(4x+4)(x^2+2x-3)^(2/3) dx

Answer 3

come on, man! It's still

∫2u^(2/3) du
= 2 (3/5) u^(5/3) + C
= 6/5 (x^2+2x-3)^(5/3) + C