# Thread: help with another integral

1. ## help with another integral

I just need help seeing part of how this is solved. These are my notes from class. The parts I do not understand are in red.

$\int(4x+3)\sqrt[3]{2x-5}$

u=2x-5
du=2dx
1/2du=dx
2x=u+5
2u+10=4x
2u+13=4x+3

I do not understand how we we concluded that if 2x=u+5 that 2u+10=4x and 2u+13 = 4x+3.

2. Originally Posted by gammaman
I just need help seeing part of how this is solved. These are my notes from class. The parts I do not understand are in red.

$\int(4x+3)\sqrt[3]{2x-5}$

u=2x-5
du=2dx
1/2du=dx
2x=u+5
2u+10=4x
2u+13=4x+3

I do not understand how we we concluded that if 2x=u+5 that 2u+10=4x and 2u+13 = 4x+3.

$2x=u+5$ multiply both sides of the equation by 2

$4x=2u+10$ add 3 to both sides

$4x+3=2u+13$ yay