I with an integral 2x dx /(x^2+y^2)^3/2

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SUMMARY

The discussion focuses on solving the integral \(\int \frac{2x \, dx}{(x^{2} + y^{2})^{3/2}}\) and successfully transforming it into \(\frac{-2}{(x^{2} + y^{2})^{1/2}}\) using the substitution method. The participant utilized the substitution \(u = x^{2} + y^{2}\), leading to \(dz = 2x \, dx\), which simplified the integral to \(\int z^{-3/2} \, dz\). This approach effectively resolved the integral, demonstrating the utility of substitution in calculus.

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Homework Statement



I need help with an integral, since my calculus skills aren't the greatest. I need help with getting from this

\pik\sigmay\int \frac{2x dx}{(x^{2} + y ^{2})^{3/2}} (i)

to this

\pik\sigmay\frac{-2}{(x^{2} + y ^{2})^{1/2}} (ii)

I integrate from 0 to a (didn't know how to get the limits into TeX in (i) and the gargantuan brackets going on either side in (ii).

Homework Equations





The Attempt at a Solution


Am I supposed to do a substitution here or? In my physics textbook I saw the numerator of (i) written as d(x^2) and the denumerator unchanged as the following step. Now, I believe I have two functions here that i need to integrate 2x and 1/(x^2+y^2)^3/2.
 
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From the looks of it, I think they have y2 as constant. So all you need to do is make a substitution like u=x2+y2.
 
This is straightforward when u notice that (x^2+y^2)'=2x. So just substitute z=x^2+y^2 and u ll have dz=2xdx and the integral becomes

\int z^{-3/2} dz
 
Thank you for your replies. I actually managed to solve the integral, but haven't been able to log on to the forum earlier. I did the u substitution and it worked fine fine. Thank you all again for your help!
 

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