anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
what's the question?
anonymous
  • anonymous
\[\int\limits_{}^{}3/(x^3-1)dx\]
anonymous
  • anonymous
is the '/' on purpose?

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anonymous
  • anonymous
Yeah, thats division.
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
I get lost when I'm at log|x-1|-\[\int\limits_{}^{}\frac{ x+2 }{ x^2+x+1}\]
anonymous
  • anonymous
ln|x-1|*
jtvatsim
  • jtvatsim
So you used partial fractions to get to this point correct?
anonymous
  • anonymous
Yeah.
jtvatsim
  • jtvatsim
Alright, so let me double check that on my end.
anonymous
  • anonymous
jtvatsim
  • jtvatsim
Cool. I checked over here and it looks correct so far.
jtvatsim
  • jtvatsim
Still thinking here... obviously, this is a tricky one. :)
amistre64
  • amistre64
what part is losing you?
amistre64
  • amistre64
\[\frac{ x+2 }{ x^2+x+1}\] \[\frac{ (x+1)+1 }{ x^2+x+1}\] what is the derivative of the bottom?
anonymous
  • anonymous
Let u=x^2+x+1 and du=2x+1dx, \[\frac{ x+2 }{ x^2+x+1 }=\frac{ 1 }{ 2 }*\frac{ 2x+1 }{ x^2+x+1 }+\frac{ 3 }{ 2 }*\frac{ 1 }{ x^2+x+1 }\]
anonymous
  • anonymous
I've gotten about up to that point, so you can say \[\int\limits_{}^{}\frac{ 1 }{ 2 }*\frac{ du }{ u }=\frac{ 1 }{ 2 }\ln|u|+C=\frac{ 1 }{ 2 }\ln|x^2+x+1|+C\]
amistre64
  • amistre64
fine, we can do that too :) \[\frac12~\frac{ 2(x+1) }{ x^2+x+1}\] \[\frac12~\frac{ 2x+2) }{ x^2+x+1}\] \[\frac12~\frac{ (2x+1)+1 }{ x^2+x+1}\]
anonymous
  • anonymous
But then \[\frac{ 3 }{ 2 } \int\limits_{}^{}\frac{ 1 }{ x^2+x+1 }dx\]
amistre64
  • amistre64
completing the square on the bottom might be useful ... been awhile tho
amistre64
  • amistre64
the other option is to just decompose it over the complex plane
anonymous
  • 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.
amistre64
  • amistre64
what are your inverse trig derivatives? reviewing them might help out ... tan^-1 rings a bell
anonymous
  • anonymous
Dont wanna type them all out so :http://tutorial.math.lamar.edu/Classes/CalcI/DiffInvTrigFcns_files/eq0044M.gif
amistre64
  • 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)
amistre64
  • amistre64
typing them out helps to keep them in memory .... pauls site wont always be available :)
anonymous
  • anonymous
True :P
anonymous
  • anonymous
\[\int\limits_{}^{}\frac{ 1 }{ (x+.5)^2+.75 }dx=\int\limits_{}^{}\frac{ \sqrt3/2*\sec^2(\theta) }{ 3/4*\tan^2\theta+1 }\]
anonymous
  • anonymous
I think.... ._.
amistre64
  • 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
anonymous
  • anonymous
Omg, \[\tan^2\theta+1=\sec^2\theta, so \it simplifies \to \int\limits_{}^{}\frac{ 2 }{ \sqrt3 }d \theta\]
amistre64
  • amistre64
you are on the right track yes http://www.wolframalpha.com/input/?i=integrate+1%2F%28x^2%2Bx%2B1%29+dx
amistre64
  • amistre64
mine went someplace off the wall lol
amistre64
  • amistre64
4/3 sqrts to get inside the ^2 ...
amistre64
  • amistre64
2/sqrt(3) not 16/9
amistre64
  • amistre64
A = 2/sqrt(3) 2K/sqrt(3) = 4/3 2sqrt(3) K = 4 K = 2/sqrt(3) thats better on my end
anonymous
  • anonymous
\[= \frac{ 2 }{ \sqrt3 } \theta+C, \theta=\arctan(\frac{ 2 }{ \sqrt3 }(x+.5)) \implies \frac{ 2 }{ \sqrt3 }\arctan(\frac{ 2 }{ \sqrt3}(x+.5))+C=\int\limits_{}^{}\frac{ 1 }{ x^2+x+1 }dx\]
amistre64
  • amistre64
now me and the wolf agree: y = K tan^-1 (A(x+B)) \[y=\frac{2}{\sqrt3}~\tan^{-1}[\frac{2}{\sqrt3}(x+\frac12)]\]
anonymous
  • anonymous
This answer is going to be really ugly in the end
anonymous
  • 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
amistre64
  • amistre64
the uglier the better :)
amistre64
  • amistre64
you got it
anonymous
  • anonymous
Jeez, that was tough.
amistre64
  • amistre64
nah .. it was "interesting" lol
anonymous
  • anonymous
Lol
amistre64
  • amistre64
good luck, its supper time
anonymous
  • anonymous
Thank you btw.

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