# Help with integration

1. Oct 2, 2008

### protivakid

Alright so I have a math problem that I have broken down into parts and one of my parts I am having trouble with is as follows.

1. The problem statement, all variables and given/known data
$$\oint$$(x2dx)/(x$$\sqrt{x2-1}$$

2. Relevant equations

3. The attempt at a solution
I attempted to solve it by making z=x2-1 and dz=2xdx. From here I am just no sure how to go about breaking this part up into something I can use. Any steers into the right direction? Thanks guys, your help really is greatly appreciated.

2. Oct 3, 2008

### jhicks

I assume you didn't mean to use a contour integral. Is this the equation you meant?

$$\int{\frac{x}{\sqrt{x^{2}-1}}dx}$$ It should be a simple substitution problem as you were doing.

3. Oct 3, 2008

### protivakid

yes that is what I meant except outside of the $$\sqrt{}$$ there is another x , that substitution is part of a much larger master problem but I have the rest figured out, I just need a helpful push into the right direction as far as solving this piece, thank you :).

4. Oct 3, 2008

### jhicks

I took one x out of the numerator and one out of the denominator. Are you sure what I posted isn't what you meant because that's all I did to simplify it algebraically.

Last edited: Oct 3, 2008
5. Oct 3, 2008

### protivakid

tuche haha still any help ?

6. Oct 3, 2008

### jhicks

Well you have a simple integral after you do your z substitution. $$\frac{1}{2}\int{\frac{dz}{\sqrt{z}}dz}=?$$

7. Oct 3, 2008

### Gib Z

I think you meant

$$\frac{1}{2} \int \frac{ \frac{dz}{dx}}{\sqrt{z}} dx$$.

8. Oct 3, 2008

### protivakid

so after putting the substitution back in I woild have 1/2$$\int$$2x/$$\sqrt{x2-1}$$ correct?

9. Oct 3, 2008

### HallsofIvy

Staff Emeritus
??
That is exactly the same as the
$$\frac{1}{2} \int\frac{dz}{\sqrt{z}}$$

That is not what you originally posted! You originally posted
$$\int \frac{x^2 dx}{x\sqrt{x^2- 1}}$$
Which is simply the same as
$$\int \frac{x dx}{\sqrt{x^2- 1}}$$
because you can cancel an "x" in numerator and denominator. Now, let z= x2- 1.

10. Oct 3, 2008

### Gib Z

Yes it is, but different to: