Question:

$\displaystyle \int(x^2-4x+4)^\frac{4}{3} . dx $

i know that $\displaystyle x^2-4x+4 $ simplifies to $\displaystyle (x-2)^2 $

So you can say the equation is:

$\displaystyle \int (x-2)^\frac{8}{3} .dx $

But i dont know how to solve it.

Please answer in steps Thank you