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.

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

Homework Helper
Gold Member
I would try writing $x^2+a^2-\sqrt 2 ax = x^2- \sqrt 2 ax+ \frac{a^2} 2 +(a^2-\frac {a^2} 2) =(x-\frac a {\sqrt2})^2+\frac{a^2} 2$ and see if that leads anywhere.

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

Last edited:
Homework Helper
I would try writing $x^2+a^2-\sqrt 2 ax = x^2-2\sqrt a + a +(a^2-a) =(x-\sqrt a)^2)+(a^2-a)$ and see if that leads anywhere.

 I see Dick types faster than I do...

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

Staff Emeritus
Homework Helper
Gold Member
I would try writing $x^2+a^2-\sqrt 2 ax = x^2-2\sqrt a + a +(a^2-a) =(x-\sqrt a)^2)+(a^2-a)$ and see if that leads anywhere.

 I see Dick types faster than I do...

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

Homework Helper
Gold Member
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.

Staff Emeritus
Homework Helper
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