Question

How do you integrate `(5x-3)^2dx`?

Answer 1

`int (5x-3)^2dx=1/15*(5x-3)^3+C`

Explanation:

Making the substitution
`t=5x-3` then `dx=1/5dt`
then we have
`1/5*int t^2dt`
using that `int x^ndx=x^(n+1)/(n+1)+C` with `n ne -1` then we obatin
`int (5x-3)^2dx=1/15*(5x-3)^3+C`