# Integral with radical denominator.

## 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.

## Answers and Replies

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

LCKurtz
Science Advisor
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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:
Dick
Science Advisor
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.

SammyS
Staff Emeritus
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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 $2(\sqrt{a})x\,.$

LCKurtz
Science Advisor
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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.

SteamKing
Staff Emeritus
Science Advisor
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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.

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 :)

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