# Homework Help: Integration by substitution

1. Jul 10, 2010

### Daveami

Hi there,

I am having difficulty with one aspect of intergration by substitution where the substituion of a square root is U^2, wondering if anyone can help.

Problem:

Integral of: 2x√(3x-4) dx by substituting U^2 = 3x-4

Would du^2/dx = 3 therefore 1/3 du^2 = dx (I think this is where im going wrong)

Im coming out with an answer of: 2/45(9x+8)(3x-4)^3/2 + k

However the answer in the book is: 4/135(9x+8)(3x-4)^3/2 + k

Any help would be greatly appreciated.

Regards

Dave

2. Jul 10, 2010

### Karmalo

Your error is that the derivative of u^2 isn't du^2.

$$u^{2}=3x-4$$

$$2u du=3dx$$

$$du=\frac{3}{2\sqrt{3x-4}}dx$$

Last edited: Jul 10, 2010
3. Jul 10, 2010

### Daveami

Ah brilliant! Thanks for the help mate!

4. Jul 10, 2010

### Staff: Mentor

And really, the substitution is u = sqrt(3x - 4). If u >= 0, this is equivalent to u^2 = 3x - 4.