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missylulu
 3 years ago
please please please help!!! double integral of 2/(x+y)^2 dydx, 2< x < 3, 1<y <5
missylulu
 3 years ago
please please please help!!! double integral of 2/(x+y)^2 dydx, 2< x < 3, 1<y <5

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KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.3First 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?

missylulu
 3 years ago
Best ResponseYou've already chosen the best response.0is it 2ln(x+5) 2ln(x+1) ???

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.3I 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 \[\int\limits_1^5 \frac{2}{(x+y)^2} dy=\left[\frac{2}{x+y}\right]^5_1=\frac{2}{x+5}\left(\frac{2}{x+1}\right)\]

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.3Does that make sense? Do you see how I did the integral?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.3If 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.

missylulu
 3 years ago
Best ResponseYou've already chosen the best response.0okok is itpossible to do it by usub?

missylulu
 3 years ago
Best ResponseYou've already chosen the best response.0will you ever use integration by parts for double integrals?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.3If you did \(u=x+3\), then you would get \((u+2)(u2)=u^24\) 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.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.3And 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.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Hmm 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?\[\large \frac{2}{x+1}\frac{2}{x+5}\]

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.0i agree with @zepdrix . the resulting expression is ripe for integration.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.3Oops. @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.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.3And this step is where the natural logs start to come in.

missylulu
 3 years ago
Best ResponseYou've already chosen the best response.0thanks guys. is the answer... 2(ln8 + ln4 +ln7 +ln3) @KingGeorge :(

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.3Hmm... I'm getting something a little different. If you had a \(\ln(3)\) at the end, that's what I'm getting.

missylulu
 3 years ago
Best ResponseYou've already chosen the best response.02(ln4  ln3  ln8 + ln7)!?

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Huge \color{blue} \checkmark \]

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.0if you simplify further, \(2(\ln 7  \ln 6)\)

missylulu
 3 years ago
Best ResponseYou've already chosen the best response.0Ok!! THANK YOU!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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