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

1. 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?

2. missylulu

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

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

4. KingGeorge

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

5. missylulu

yes!!

6. 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.

7. missylulu

okok is itpossible to do it by usub?

8. missylulu

will you ever use integration by parts for double integrals?

9. 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.

10. 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.

11. 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?$\large \frac{2}{x+1}-\frac{2}{x+5}$

12. sirm3d

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

13. 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.

14. KingGeorge

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

15. missylulu

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

16. KingGeorge

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

17. missylulu

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

18. sirm3d

$\Huge \color{blue} \checkmark$

19. sirm3d

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

20. missylulu

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

21. KingGeorge

Looks great.