## anonymous one year ago Okay, so if you don't like math then don't try to answer this. I've spent the last hour doing it. ∫3/(x^3−1)dx

1. anonymous

what's the question?

2. anonymous

$\int\limits_{}^{}3/(x^3-1)dx$

3. anonymous

is the '/' on purpose?

4. anonymous

Yeah, thats division.

5. anonymous

oh ok

6. anonymous

I get lost when I'm at log|x-1|-$\int\limits_{}^{}\frac{ x+2 }{ x^2+x+1}$

7. anonymous

ln|x-1|*

8. jtvatsim

So you used partial fractions to get to this point correct?

9. anonymous

Yeah.

10. jtvatsim

Alright, so let me double check that on my end.

11. anonymous

I got

12. jtvatsim

Cool. I checked over here and it looks correct so far.

13. jtvatsim

Still thinking here... obviously, this is a tricky one. :)

14. amistre64

what part is losing you?

15. amistre64

$\frac{ x+2 }{ x^2+x+1}$ $\frac{ (x+1)+1 }{ x^2+x+1}$ what is the derivative of the bottom?

16. anonymous

Let u=x^2+x+1 and du=2x+1dx, $\frac{ x+2 }{ x^2+x+1 }=\frac{ 1 }{ 2 }*\frac{ 2x+1 }{ x^2+x+1 }+\frac{ 3 }{ 2 }*\frac{ 1 }{ x^2+x+1 }$

17. anonymous

I've gotten about up to that point, so you can say $\int\limits_{}^{}\frac{ 1 }{ 2 }*\frac{ du }{ u }=\frac{ 1 }{ 2 }\ln|u|+C=\frac{ 1 }{ 2 }\ln|x^2+x+1|+C$

18. amistre64

fine, we can do that too :) $\frac12~\frac{ 2(x+1) }{ x^2+x+1}$ $\frac12~\frac{ 2x+2) }{ x^2+x+1}$ $\frac12~\frac{ (2x+1)+1 }{ x^2+x+1}$

19. anonymous

But then $\frac{ 3 }{ 2 } \int\limits_{}^{}\frac{ 1 }{ x^2+x+1 }dx$

20. amistre64

completing the square on the bottom might be useful ... been awhile tho

21. amistre64

the other option is to just decompose it over the complex plane

22. anonymous

Oh okay, so its $\frac{ 1 }{ (x+.5)^2+.75 }$ ? and now I'm pretty lost. It looks like maybe you can do some trig substitution but I can't really tell.

23. amistre64

what are your inverse trig derivatives? reviewing them might help out ... tan^-1 rings a bell

24. anonymous

Dont wanna type them all out so : http://tutorial.math.lamar.edu/Classes/CalcI/DiffInvTrigFcns_files/eq0044M.gif

25. amistre64

y = tan^-1 (x) tan(y) = x y' sec^2(y) = 1 y' = 1/sec^2(y) y' = 1/(tan^2(y)+1) y' = 1/(tan^2(tan^-1(x))+1) y' = 1/(x^2+1)

26. amistre64

typing them out helps to keep them in memory .... pauls site wont always be available :)

27. anonymous

True :P

28. anonymous

$\int\limits_{}^{}\frac{ 1 }{ (x+.5)^2+.75 }dx=\int\limits_{}^{}\frac{ \sqrt3/2*\sec^2(\theta) }{ 3/4*\tan^2\theta+1 }$

29. anonymous

I think.... ._.

30. amistre64

im pretty sure im making a mess lol, but heres my thought process if we go this archaic route 1 -------------- (x+1/2)^2 + 3/4 4/3 ---------------- 4/3(x+1/2)^2 + 1 4/3 ---------------- (16/9 (x+1/2))^2 + 1 assuming we can work out some tan inverse y = K tan^-1(A(x+B)) y/K = tan^-1(A(x+B)) tan(y/K) = A(x+B) y' sec^2(y/K)/K = A y' sec^2(y/K) = KA y' (tan^2(y/K)+1) = KA y' ((A(x+B))^2+1) = KA y' = KA ------------ (A(x+B))^2+1 ------------------- A = 16/9, B=1/2 16K/9 = 4/3 16K = 12 K = 3/4

31. anonymous

Omg, $\tan^2\theta+1=\sec^2\theta, so \it simplifies \to \int\limits_{}^{}\frac{ 2 }{ \sqrt3 }d \theta$

32. amistre64

you are on the right track yes http://www.wolframalpha.com/input/?i=integrate+1%2F%28x^2%2Bx%2B1%29+dx

33. amistre64

mine went someplace off the wall lol

34. amistre64

4/3 sqrts to get inside the ^2 ...

35. amistre64

2/sqrt(3) not 16/9

36. amistre64

A = 2/sqrt(3) 2K/sqrt(3) = 4/3 2sqrt(3) K = 4 K = 2/sqrt(3) thats better on my end

37. anonymous

$= \frac{ 2 }{ \sqrt3 } \theta+C, \theta=\arctan(\frac{ 2 }{ \sqrt3 }(x+.5)) \implies \frac{ 2 }{ \sqrt3 }\arctan(\frac{ 2 }{ \sqrt3}(x+.5))+C=\int\limits_{}^{}\frac{ 1 }{ x^2+x+1 }dx$

38. amistre64

now me and the wolf agree: y = K tan^-1 (A(x+B)) $y=\frac{2}{\sqrt3}~\tan^{-1}[\frac{2}{\sqrt3}(x+\frac12)]$

39. anonymous

This answer is going to be really ugly in the end

40. anonymous

Yeah, I agree, then we just put it all together and its $\log|x-1|-0.5\log|x^2+x+1|-\sqrt3\arctan(\frac{ 2 }{ \sqrt3 }(x+.5)+C$ I dont think that simplifies any further

41. amistre64

the uglier the better :)

42. amistre64

you got it

43. anonymous

Jeez, that was tough.

44. amistre64

nah .. it was "interesting" lol

45. anonymous

Lol

46. amistre64

good luck, its supper time

47. anonymous

Thank you btw.