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

[crossroads]

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.

 

[tiny-tim]

Welcome to PF!

Hi crossroads! Welcome to PF!

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

Yes, x² is a constant.

No, it isn't -1/y√(x² + y²).

Hint: try a trig substitution. ​

 

[crossroads]

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.