Solving U-Substitution Problem with 2x√(2x-3) in Calculus

[jdawg]

Homework Statement

∫2x√(2x-3) dx

Homework Equations

The Attempt at a Solution

u=2x
du=2 dx

1/2∫u√(u-3) du

Am I on the right track with this? I'm not really sure what to do next.

 

[Jtechguy21]

Sort of. u= the inside of the square root
u=2x-3
du=2dx
dx=du/2 or 1/2

∫1/2 2xu^1/2 dx
the 2's cancel out.
now it should look like
∫x*u^1/2 and integrate that. and plug in the value for u after your done integrating
Becareful, because if i remember correctly. after integrating ∫x*u^1/2
Since we have an x. you have to solve for it, and plug it in.
u=2x-3 <-Solve the x

2x-3=0 2x=3
x=3/2

 

[jdawg]

Thanks so much!

 

[SteamKing]

Sort of. u= the inside of the square root
u=2x-3
du=2dx
dx=du/2 or 1/2

∫1/2 2xu^1/2 dx
the 2's cancel out.
now it should look like
∫x*u^1/2 and integrate that. and plug in the value for u after your done integrating
Becareful, because if i remember correctly. after integrating ∫x*u^1/2
Since we have an x. you have to solve for it, and plug it in.
u=2x-3 <-Solve the x

2x-3=0 2x=3
x=3/2

If u = 2x -3, then 2x = u + 3, so that your integral after substitution has only u in it.

Your integrand becomes (u+3)*SQRT(u)*du/2

You don't want an integrand which mixes x and u after substitution.

 

[Jtechguy21]

If u = 2x -3, then 2x = u + 3, so that your integral after substitution has only u in it.

Your integrand becomes (u+3)*SQRT(u)*du/2

You don't want an integrand which mixes x and u after substitution.

thanks for the correction