Question

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

Answer 1

what's the question?

Answer 2

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

Answer 3

is the '/' on purpose?

Answer 4

Yeah, thats division.

Answer 5

oh ok

Answer 6

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

Answer 7

ln|x-1|*

Answer 8

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

Answer 9

Yeah.

Answer 10

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

Answer 11

I got

Answer 12

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

Answer 13

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

Answer 14

what part is losing you?

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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.

Answer 23

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

Answer 24

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

Answer 25

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)

Answer 26

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

Answer 27

True :P

Answer 28

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

Answer 29

I think.... ._.

Answer 30

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

Answer 31

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

Answer 32

you are on the right track yes

http://www.wolframalpha.com/input/?i=integrate+1%2F%28x^2%2Bx%2B1%29+dx

Answer 33

mine went someplace off the wall lol

Answer 34

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

Answer 35

2/sqrt(3) not 16/9

Answer 36

A = 2/sqrt(3)

2K/sqrt(3) = 4/3

2sqrt(3) K = 4

K = 2/sqrt(3)

thats better on my end

Answer 37

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

Answer 38

now me and the wolf agree:

y = K tan^-1 (A(x+B))

\[y=\frac{2}{\sqrt3}~\tan^{-1}[\frac{2}{\sqrt3}(x+\frac12)]\]

Answer 39

This answer is going to be really ugly in the end

Answer 40

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

Answer 41

the uglier the better :)

Answer 42

you got it

Answer 43

Jeez, that was tough.

Answer 44

nah .. it was "interesting" lol

Answer 45

Lol

Answer 46

good luck, its supper time

Answer 47

Thank you btw.