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

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KingGeorgeBest ResponseYou've already chosen the best response.3
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?
 one year ago

missyluluBest ResponseYou've already chosen the best response.0
is it 2ln(x+5) 2ln(x+1) ???
 one year ago

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

KingGeorgeBest ResponseYou've already chosen the best response.3
Does that make sense? Do you see how I did the integral?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.3
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.
 one year ago

missyluluBest ResponseYou've already chosen the best response.0
okok is itpossible to do it by usub?
 one year ago

missyluluBest ResponseYou've already chosen the best response.0
will you ever use integration by parts for double integrals?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.3
If 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.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.3
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.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
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?\[\large \frac{2}{x+1}\frac{2}{x+5}\]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.0
i agree with @zepdrix . the resulting expression is ripe for integration.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.3
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.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.3
And this step is where the natural logs start to come in.
 one year ago

missyluluBest ResponseYou've already chosen the best response.0
thanks guys. is the answer... 2(ln8 + ln4 +ln7 +ln3) @KingGeorge :(
 one year ago

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

missyluluBest ResponseYou've already chosen the best response.0
2(ln4  ln3  ln8 + ln7)!?
 one year ago

sirm3dBest ResponseYou've already chosen the best response.0
\[\Huge \color{blue} \checkmark \]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.0
if you simplify further, \(2(\ln 7  \ln 6)\)
 one year ago

missyluluBest ResponseYou've already chosen the best response.0
Ok!! THANK YOU!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
 one year ago
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