How Do You Evaluate the Integral of (x^2 + y^2)^(-3/2) with Respect to y?

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To evaluate the integral of (x^2 + y^2)^(-3/2) with respect to y, it is confirmed that x^2 can be treated as a constant. A suggestion is made to use trigonometric substitution to simplify the process. The initial attempt at integration yields an incorrect result, indicating a need for further review of integration techniques. The participant successfully arrives at the correct answer after revisiting trigonometric substitutions. This discussion highlights the importance of understanding integration methods for solving complex integrals.
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I took calc a year ago, and already am apparently rusty. I'm trying to figure out how to evaluate the integral: \int (x^2+y^2)^(^-^3^/^2) dy. This isn't a homework problem by the way, its actually an example problem in the book, but they don't show step by step how they got to the answer. If someone could go through this step by step that would be really nice :). You're just supposed to treat the x^2 as a constant, right? I get as far as: (-2 (x^2+y^2)^(^-^1^/^2^)) / 2y but I'm not even sure if that much is correct. Thanks for any help.
 
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Welcome to PF!

Hi crossroads! Welcome to PF! :smile:

(have an integral: ∫ and a square-root: √ and try using the X2 tag just above the Reply box :wink:)

Yes, x2 is a constant.

No, it isn't -1/y√(x2 + y2).

Hint: try a trig substitution. :wink:​
 
Thanks for the hint and the welcome. I got the answer, but it looks like I need to review my trig substitutions and integration by substitution.
 

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