Get Step-by-Step Help with Integration for a Tricky Math Problem

[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.

Homework Statement

\oint(x²dx)/(x\sqrt{x²-1}

Homework Equations

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.

 

[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.

 

[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 :).

 

[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.

 

[protivakid]

tuche haha still any help ?

 

[jhicks]

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

 

[Gib Z]

I think you meant

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

 

[protivakid]

so after putting the substitution back in I woild have 1/2\int2x/\sqrt{x²-1} correct?

 

[HallsofIvy]

I think you meant

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

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

so after putting the substitution back in I woild have 1/2\int2x/\sqrt{x²-1} correct?

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.

 

[Gib Z]

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

Yes it is, but different to:

. \frac{1}{2}\int{\frac{dz}{\sqrt{z}}dz}=?