1. ## Intergrals

The intergral of (4-2x)^3 dx?

2. substitute $u=4-2x$

3. Originally Posted by homerb
The intergral of (4-2x)^3 dx?
Use substitution:

Let:   $u=4-2x, \text{ then } du=-2\,dx$

4. thats what I did and I got -1/8*(4-2x)^4 + c, but in the book says -2(x-4)^4, am I doing something wrong?

$-\dfrac{1}{8} (4-2x)^4 + C = -\dfrac{1}{8} [-2(x-2)]^4 + C = ...$