## anonymous 5 years ago what is the integration of 1/{(x+2)(sqr of x+3)} ?

1. anonymous

Is that suppose to be 1/{(x+2)(x+3)(x+3)} What is squared?

2. anonymous

no...it is $1/({(x+2)\sqrt{x+3}})$

3. anonymous

how can this be solved?

4. anonymous

What Calculus class are you in?

5. anonymous

this can be solved using u subs?

6. anonymous

What Calculus class are you in?

7. anonymous

what?

8. anonymous

Are you doing this for fun or a class?

9. anonymous

class of course...wanted to learn

10. anonymous

What type of math class? Clause it depends on what way.

11. anonymous

12. anonymous

"learned" so you're in basic calculus, trying to get ahead?

13. anonymous

yeah

14. anonymous

OK, what ways do you know how to use?

15. anonymous

for this prob, i guess we can use integration by parts..right? but i always get confused in which term to take as 'u' and which as 'v'....i need some hints..

16. anonymous

In order to do that your going to need to change some parts and move them around a little. I would change the square root to a power and take it out from the demomatator.

17. anonymous

like denominator=(x+2).(x+3)$^{1/2}$

18. anonymous

(x+2).(x+3)1/2

19. anonymous

with the 1/2 on the (x+3) yes then pull it out so it becomes a product rule looking equation

20. anonymous

then it would look like $1\div ((x+2).(x+3)^{1/2})$

21. anonymous

$1\div \left( x+2 \right)\left( x+3 \right)^{1/2}$ to $\left( x+3 \right)^{-1/2}$ next to 1÷(x+2)

22. anonymous

okay so now it is $\int\limits_{}^{} [(x+3)^{-1/2}]/(x+2)$

23. anonymous

you could look at it that way but it might be more helpful if you had that in two separate parts. might be easier to work with

24. anonymous

Think product rule versus quot. rule

25. anonymous

so can you solve this now?..i am solving it right now...plz do solve and let me know the answer..

26. anonymous

This has taken longer than I planned, and I'm about to be sleeping on my keyboard. I hope this helps

27. anonymous

correct me if i am wrong , but after intr. for the first time i get, 2$\sqrt{x+3}/(x+2)-(2 \int\limits_{}^{}\sqrt{x+3} . \log(x+3))dx$

28. anonymous

I think your missing a /2 in the first part, but my brain is turning off. night, sorry but I'm just too tired to think much more, worst case I'll check back tomorrow.

29. anonymous

okay....i'll try by then....

30. anonymous

did u get the solution?

31. anonymous

yes, I'm going to type the whole solution out for you... hold on

32. anonymous

thanks

33. anonymous

$\int\limits \frac{1}{(x+2)\sqrt {x+3}}dx \rightarrow u=\sqrt{x+3}, 2du=\frac{1}{\sqrt{x+3}}dx$ $2 \int\limits \frac{1}{x+2}du \rightarrow now:u^2-1=x+2, so \rightarrow 2 \int\limits \frac{1}{u^2-1}du$ Use partial fraction decomposition: $\frac{1}{u^2-1}=\frac {1}{(u+1)(u-1)}= \frac{A}{u+1}+\frac {B}{u-1}$ $A(u-1)+B(u+1)=1\rightarrow Au-A+Bu+B=1$ $(B+A)u=0u, B-A=1\rightarrow so: B+A=0, B-A=1$ solve for B and A: $B=\frac{1}{2}, A=- \frac{1}{2}$ $2 \int\limits\limits \frac{1}{2} (\frac{1}{u-1}-\frac{1}{u+1})du \rightarrow \int\limits \frac{1}{u-1}du- \int\limits \frac{1}{u+1}du$ $\int\limits\limits \frac{1}{u-1}du \rightarrow v=u-1, dv=du$ $\rightarrow \int\limits\limits \frac{1}{v}dv=\ln(v)=\ln(u-1)$ $\int\limits \frac {1}{u+1}du \rightarrow z=u+1, dz=du$ $\rightarrow \int\limits \frac{1}{z}dz=\ln(z)=\ln(u+1)$ so:$\ln(u-1)-\ln(u+1)+C \rightarrow \ln \left| \frac {u-1}{u+1} \right|+C$ From the begining we know what u is, so: $since:u=\sqrt{x+3}\rightarrow \ln \left| \frac{\sqrt{x+3}-1}{\sqrt{x+3}+1}\right|+C$

34. anonymous

how did you take : u2−1=x+2 ?

35. anonymous

like this: $u=\sqrt{x+3}\rightarrow u^2=(\sqrt{x+3})^2\rightarrow u^2=x+3\rightarrow u^2-1=x+2$

36. anonymous

37. anonymous

there's one more question...if u can answer...its regarding partial derivatives..can u?

38. anonymous

sure

39. anonymous

its just 2 places above this post.. which is like fxy =(-y sin xy)...so on...plz move to that post...

40. anonymous

i got it :)