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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
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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what's the question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}3/(x^31)dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is the '/' on purpose?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, thats division.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I get lost when I'm at logx1\[\int\limits_{}^{}\frac{ x+2 }{ x^2+x+1}\]

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.0So you used partial fractions to get to this point correct?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.0Alright, so let me double check that on my end.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.0Cool. I checked over here and it looks correct so far.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.0Still thinking here... obviously, this is a tricky one. :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1what part is losing you?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{ x+2 }{ x^2+x+1}\] \[\frac{ (x+1)+1 }{ x^2+x+1}\] what is the derivative of the bottom?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let 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
 one year ago
Best ResponseYou've already chosen the best response.0I've gotten about up to that point, so you can say \[\int\limits_{}^{}\frac{ 1 }{ 2 }*\frac{ du }{ u }=\frac{ 1 }{ 2 }\lnu+C=\frac{ 1 }{ 2 }\lnx^2+x+1+C\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1fine, 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
 one year ago
Best ResponseYou've already chosen the best response.0But then \[\frac{ 3 }{ 2 } \int\limits_{}^{}\frac{ 1 }{ x^2+x+1 }dx\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1completing the square on the bottom might be useful ... been awhile tho

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the other option is to just decompose it over the complex plane

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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
 one year ago
Best ResponseYou've already chosen the best response.1what are your inverse trig derivatives? reviewing them might help out ... tan^1 rings a bell

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Dont wanna type them all out so : http://tutorial.math.lamar.edu/Classes/CalcI/DiffInvTrigFcns_files/eq0044M.gif

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1y = 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
 one year ago
Best ResponseYou've already chosen the best response.1typing them out helps to keep them in memory .... pauls site wont always be available :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}\frac{ 1 }{ (x+.5)^2+.75 }dx=\int\limits_{}^{}\frac{ \sqrt3/2*\sec^2(\theta) }{ 3/4*\tan^2\theta+1 }\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1im 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
 one year ago
Best ResponseYou've already chosen the best response.0Omg, \[\tan^2\theta+1=\sec^2\theta, so \it simplifies \to \int\limits_{}^{}\frac{ 2 }{ \sqrt3 }d \theta\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1you are on the right track yes http://www.wolframalpha.com/input/?i=integrate+1%2F%28x^2%2Bx%2B1%29+dx

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1mine went someplace off the wall lol

amistre64
 one year ago
Best ResponseYou've already chosen the best response.14/3 sqrts to get inside the ^2 ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1A = 2/sqrt(3) 2K/sqrt(3) = 4/3 2sqrt(3) K = 4 K = 2/sqrt(3) thats better on my end

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[= \frac{ 2 }{ \sqrt3 } \theta+C, \theta=\arctan(\frac{ 2 }{ \sqrt3 }(x+.5)) \implies \frac{ 2 }{ \sqrt3 }\arctan(\frac{ 2 }{ \sqrt3}(x+.5))+C=\int\limits_{}^{}\frac{ 1 }{ x^2+x+1 }dx\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1now me and the wolf agree: y = K tan^1 (A(x+B)) \[y=\frac{2}{\sqrt3}~\tan^{1}[\frac{2}{\sqrt3}(x+\frac12)]\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This answer is going to be really ugly in the end

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, I agree, then we just put it all together and its \[\logx10.5\logx^2+x+1\sqrt3\arctan(\frac{ 2 }{ \sqrt3 }(x+.5)+C\] I dont think that simplifies any further

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the uglier the better :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Jeez, that was tough.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1nah .. it was "interesting" lol

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1good luck, its supper time
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