Integral with radical denominator.

In summary, The conversation revolved around solving a specific integral, \int \frac{xdx}{\sqrt{x^2+a^2-\sqrt{2}ax}}, where the participants were trying to determine the appropriate substitution and approach for solving it. The conversation concluded with the correct answer being \sqrt{x^2+a^2+\sqrt{2}ax} + \frac{a}{\sqrt{2}} ln |\sqrt{a^2+x^2} + x |.
  • #1
Idoubt
172
1

Homework Statement




Need to do this integral as part of a problem

[itex]\int[/itex] [itex]\frac{xdx}{\sqrt{x^2+a^2-\sqrt{2}ax}}[/itex]


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.
 
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  • #2
Your first step would be completing the square in the denominator. What does the integral look like after you do that?
 
  • #3
I would try writing [itex]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[/itex] and see if that leads anywhere.

[Edit] I see Dick types faster than I do...
[Edit] Fixed a typo and its ramifications.
 
Last edited:
  • #4
LCKurtz said:
I would try writing [itex]x^2+a^2-\sqrt 2 ax = x^2-2\sqrt a + a +(a^2-a)
=(x-\sqrt a)^2)+(a^2-a)[/itex] 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.
 
  • #5
LCKurtz said:
I would try writing [itex]x^2+a^2-\sqrt 2 ax = x^2-2\sqrt a + a +(a^2-a)
=(x-\sqrt a)^2)+(a^2-a)[/itex] and see if that leads anywhere.

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

It appears that you have changed [itex]\sqrt{2}ax[/itex] into [itex]2\sqrt{a}[/itex], but apparently meant [itex]2(\sqrt{a})x\,.[/itex]
 
  • #6
SammyS said:
It appears that you have changed [itex]\sqrt{2}ax[/itex] into [itex]2\sqrt{a}[/itex], but apparently meant [itex]2(\sqrt{a})x\,.[/itex]

Yes, thanks. I edited it (about 3 times :frown:) to correct it.
 
  • #7
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.
 
  • #8
ok when I complete the squares and do it, i get the answer as


[itex]\sqrt{x^2+a^2+\sqrt{2}ax}[/itex] + [itex]\frac{a}{\sqrt{2}}[/itex] ln |[itex]\sqrt{a^2+x^2}[/itex] + x |

Which is the right answer i think. thanks a lot for the help :)
 
  • #9
SteamKing said:
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
 

Related to Integral with radical denominator.

What is an integral with a radical denominator?

An integral with a radical denominator is a type of mathematical expression that involves finding the area under a curve where the denominator contains a radical (square root) symbol.

Why is it important to solve integrals with radical denominators?

Solving integrals with radical denominators can be useful in many areas of science and engineering, as it allows us to calculate important quantities such as the volume of a solid or the work done by a force.

What is the general method for solving integrals with radical denominators?

The general method for solving integrals with radical denominators is to first simplify the expression by using algebraic techniques, such as rationalizing the denominator. Then, we can use integration techniques such as substitution or integration by parts to evaluate the integral.

What are some common mistakes made when solving integrals with radical denominators?

One common mistake when solving integrals with radical denominators is forgetting to rationalize the denominator. It is also important to be careful with the signs of the terms in the expression, as this can affect the final result.

Are there any tips for solving integrals with radical denominators more efficiently?

One tip for solving integrals with radical denominators more efficiently is to carefully choose the substitution variable, as this can make the integration process simpler. It can also be helpful to use trigonometric identities or other algebraic techniques to simplify the expression before integrating.

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