anonymous
  • anonymous
please please please help!!! double integral of 2/(x+y)^2 dydx, 2< x < 3, 1
Mathematics
katieb
  • katieb
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KingGeorge
  • KingGeorge
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?
anonymous
  • anonymous
is it 2ln(x+5) -2ln(x+1) ???
KingGeorge
  • KingGeorge
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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KingGeorge
  • KingGeorge
Does that make sense? Do you see how I did the integral?
anonymous
  • anonymous
yes!!
KingGeorge
  • KingGeorge
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.
anonymous
  • anonymous
okok is itpossible to do it by usub?
anonymous
  • anonymous
will you ever use integration by parts for double integrals?
KingGeorge
  • KingGeorge
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.
KingGeorge
  • KingGeorge
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.
zepdrix
  • zepdrix
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?\[\large \frac{2}{x+1}-\frac{2}{x+5}\]
sirm3d
  • sirm3d
i agree with @zepdrix . the resulting expression is ripe for integration.
KingGeorge
  • KingGeorge
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.
KingGeorge
  • KingGeorge
And this step is where the natural logs start to come in.
anonymous
  • anonymous
thanks guys. is the answer... 2(-ln8 + ln4 +ln7 +ln3) @KingGeorge :(
KingGeorge
  • KingGeorge
Hmm... I'm getting something a little different. If you had a \(-\ln(3)\) at the end, that's what I'm getting.
anonymous
  • anonymous
2(ln4 - ln3 - ln8 + ln7)!?
sirm3d
  • sirm3d
\[\Huge \color{blue} \checkmark \]
sirm3d
  • sirm3d
if you simplify further, \(2(\ln 7 - \ln 6)\)
anonymous
  • anonymous
Ok!! THANK YOU!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
KingGeorge
  • KingGeorge
Looks great.

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