Question

please please please help!!! double integral of 2/(x+y)^2 dydx, 2< x < 3, 1 13 years ago

Answer 1

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?

Answer 2

is it 2ln(x+5) -2ln(x+1) ???

Answer 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 \[\int\limits_1^5 \frac{2}{(x+y)^2} dy=\left[-\frac{2}{x+y}\right]^5_1=-\frac{2}{x+5}-\left(-\frac{2}{x+1}\right)\]

Answer 4

Does that make sense? Do you see how I did the integral?

Answer 5

yes!!

Answer 6

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.

Answer 7

okok is itpossible to do it by usub?

Answer 8

will you ever use integration by parts for double integrals?

Answer 9

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.

Answer 10

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.

Answer 11

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?\[\large \frac{2}{x+1}-\frac{2}{x+5}\]

Answer 12

i agree with @zepdrix . the resulting expression is ripe for integration.

Answer 13

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.

Answer 14

And this step is where the natural logs start to come in.

Answer 15

thanks guys. is the answer... 2(-ln8 + ln4 +ln7 +ln3) @KingGeorge :(

Answer 16

Hmm... I'm getting something a little different. If you had a \(-\ln(3)\) at the end, that's what I'm getting.

Answer 17

2(ln4 - ln3 - ln8 + ln7)!?

Answer 18

\[\Huge \color{blue} \checkmark \]

Answer 19

if you simplify further, \(2(\ln 7 - \ln 6)\)

Answer 20

Ok!! THANK YOU!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 21

Looks great.