Integral with radical denominator.

[Idoubt]

Homework Statement

Need to do this integral as part of a problem

$\int$ $\frac{xdx}{\sqrt{x^2+a^2-\sqrt{2}ax}}$


I can't think of any substitution that would work, or any way to factorize.
I think this is a standard integral, but can't remember everyone so I just want to learn to solve it.

 

[Dick]

Your first step would be completing the square in the denominator. What does the integral look like after you do that?

 

[LCKurtz]

I would try writing $x^2+a^2-\sqrt 2 ax = x^2- \sqrt 2 ax+ \frac{a^2} 2 +(a^2-\frac {a^2} 2)<br /> =(x-\frac a {\sqrt2})^2+\frac{a^2} 2$ and see if that leads anywhere.

[Edit] I see Dick types faster than I do...
[Edit] Fixed a typo and its ramifications.

 

[Dick]

I would try writing $x^2+a^2-\sqrt 2 ax = x^2-2\sqrt a + a +(a^2-a)<br /> =(x-\sqrt a)^2)+(a^2-a)$ and see if that leads anywhere.

[Edit] I see Dick types faster than I do...

No, we type about the same. I just say less.

 

[SammyS]

I would try writing $x^2+a^2-\sqrt 2 ax = x^2-2\sqrt a + a +(a^2-a)<br /> =(x-\sqrt a)^2)+(a^2-a)$ and see if that leads anywhere.

[Edit] I see Dick types faster than I do...

It appears that you have changed $\sqrt{2}ax$ into $2\sqrt{a}$, but apparently meant $2(\sqrt{a})x\,.$

 

[LCKurtz]

It appears that you have changed $\sqrt{2}ax$ into $2\sqrt{a}$, but apparently meant $2(\sqrt{a})x\,.$

Yes, thanks. I edited it (about 3 times ) to correct it.

 

[SteamKing]

You are letting the radicals and the denominator distract you. If you re-write as
Int (mess)^(-1/2) * xdx, you should see a u-substitution and integration by parts.

 

[Idoubt]

ok when I complete the squares and do it, i get the answer as


$\sqrt{x^2+a^2+\sqrt{2}ax}$ + $\frac{a}{\sqrt{2}}$ ln |$\sqrt{a^2+x^2}$ + x |

Which is the right answer i think. thanks a lot for the help :)

 

[Idoubt]

You are letting the radicals and the denominator distract you. If you re-write as
Int (mess)^(-1/2) * xdx, you should see a u-substitution and integration by parts.

If i do the U substitution, I don't see how i can write x in terms of u