please please please help!!! double integral of 2/(x+y)^2 dydx, 2< x < 3, 1

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please please please help!!! double integral of 2/(x+y)^2 dydx, 2< x < 3, 1

Mathematics
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First step, you just need to find the following integral where you just pretend \(x\) is some constant. \[\int\limits_1^5 \frac{2}{(x+y)^2} dy\]Can you do this yourself?
is it 2ln(x+5) -2ln(x+1) ???
I think you kind of have the idea. That would be the integral if the denominator did not have a power of 2. So you would get \[\int\limits_1^5 \frac{2}{(x+y)^2} dy=\left[-\frac{2}{x+y}\right]^5_1=-\frac{2}{x+5}-\left(-\frac{2}{x+1}\right)\]

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Does that make sense? Do you see how I did the integral?
yes!!
If you simplify that, you should get\[-\frac{2}{x+5}-\left(-\frac{2}{x+1}\right)=\frac{8}{5+6 x+x^2}=\frac{8}{(x+5)(x+1)}\]With this, I imagine you could use partial fractions, and then you'll end up with two integrals of x which should be fairly straightforward. I'll let you work out the rest on your own. If you get stuck, let me know and I can help some more.
okok is itpossible to do it by usub?
will you ever use integration by parts for double integrals?
If you did \(u=x+3\), then you would get \((u+2)(u-2)=u^2-4\) in the denominator. I suppose you could then do something with the trig substitutions. I really think that it might be better if you do the partial fractions.
And you can do integration by parts in double integrals, but in my experience, you don't usually see it simply because you don't need it.
Hmm I'm confused, how did you end up needing to use partial fractions King? Why not integrate in X from where you left off after evaluating the upper and lower limits of y?\[\large \frac{2}{x+1}-\frac{2}{x+5}\]
i agree with @zepdrix . the resulting expression is ripe for integration.
Oops. @zepdrix is correct. We don't even need to do partial fractions. \[\int\limits_2^3\frac{2}{x+1}dx-\int\limits_2^3\frac{2}{x+5}dx\]Integrating this should work.
And this step is where the natural logs start to come in.
thanks guys. is the answer... 2(-ln8 + ln4 +ln7 +ln3) @KingGeorge :(
Hmm... I'm getting something a little different. If you had a \(-\ln(3)\) at the end, that's what I'm getting.
2(ln4 - ln3 - ln8 + ln7)!?
\[\Huge \color{blue} \checkmark \]
if you simplify further, \(2(\ln 7 - \ln 6)\)
Ok!! THANK YOU!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Looks great.

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