Question

integration: Can someone help me get started with this question please integrate
I will type it below

Answer 1

\[integrate 4x \sqrt[3](7-2x) dx{?}\]n the square root is over the entire 7-2x

Answer 2

try substituting u= 7-2x and du = -2 dx

Answer 3

I did let u =7-2x du/dx=-2 and dy=du/-2

Answer 4

just not sure what the next steps are

Answer 5

4x (7-2x)^(1/3) dx --> 4 (7-u)/2 (u^(1/3)) (1/-2) du = (u-7) u^(1/3) du

multiply through by u^1/3 du: u^(4/3) du -7 u^(1/3) du

Answer 6

with u= 7-2x we have x= (7-u)/2 and dx= du/-2 which we sub into the equation