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 November 26th, 2012, 01:05 PM #1 Member   Joined: Sep 2012 Posts: 34 Thanks: 0 Double Integral Problem Hey everyone, I am having a bit of difficulty with a particular double integral problem. It is the integral of: (x-y)/(x+y)^3 dydx ; where [x=0 to x=2] and [y=0 to y=1] How would I solve this problem with respect to y first? I've thought about using partial fractions, but I am not quite sure as to how to apply it within this particular problem. Is there perhaps another method? Any feedback and help is appreciated, thanks.
 November 26th, 2012, 01:59 PM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Double Integral Problem It really doesn't matter whether you integrate with respect to x and y first, it's clear they calculations will be basically the same. You have $\int_{x= 0}^2 \int_{y= 0}^1\frac{x- y}{(x+y)^3} dy dx$ No, you would not use "partial fractions" because you already have a linear term over the third power. Instead, make the substitution u= x+ y. The dy= du, when y= 0, u= x, when y= 1, u= x+ 1, and y= u- x so that $\frac{x- y}{(x+y)^3}= \frac{x- (u- x)}{u^3}= \frac{2x- u}{u^3}$. The integral becomes $\int_{x= 0}^2 \int_{u= x}^{x+1} 2xu^{-3}- u^{-2} du dx$.
 November 26th, 2012, 02:44 PM #3 Member   Joined: Sep 2012 Posts: 34 Thanks: 0 Re: Double Integral Problem Nice, thanks for the explanation.

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