1. KingGeorge Group Title

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 Group Title

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

3. KingGeorge Group Title

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 Group Title

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

5. missylulu Group Title

yes!!

6. KingGeorge Group Title

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 Group Title

okok is itpossible to do it by usub?

8. missylulu Group Title

will you ever use integration by parts for double integrals?

9. KingGeorge Group Title

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 Group Title

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 Group Title

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 Group Title

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

13. KingGeorge Group Title

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 Group Title

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

15. missylulu Group Title

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

16. KingGeorge Group Title

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 Group Title

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

18. sirm3d Group Title

$\Huge \color{blue} \checkmark$

19. sirm3d Group Title

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

20. missylulu Group Title

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

21. KingGeorge Group Title

Looks great.