Question

Partial Fraction indef. integral of:
(e^x)/(1-e^(3x))

tried wolframalpha, and got to the u substitution.. it got really complicated after that.. and I need help.

Thanks!

Answer 1

\[\int\limits_{}^{} \frac{ e^x }{ 1-e ^{3x} }\]

Answer 2

forgot the dx at the end of the integral, sorry

Answer 3

How would you use partial fractions to decompose this fraction:
\[\frac{u}{1-u^3}\]

Answer 4

change the sign to negative.. giving you negative integral of u/(u^3-1)

Answer 5

you should be able to factor a difference of cubes

Answer 6

making it a polynomial.. which i got.. but after solving and finding out what a, bx and c are, you have to repeat?

Answer 7

oh, yeah.. that's (u-1)(u^2 + u + 1)

Answer 8

yep
so one of those factors is linear and the other factor is an irreducible quadratic
we are done with the factoring bit

Answer 9

for linear factors, you do constant/linear
for irreducible quadratics, you do (constant*the variable +another constant)/the quadratic

Answer 10

\[\frac{-u}{u^3-1}=\frac{A}{u-1}+\frac{Bu+C}{u^2+u+1}\]

Answer 11

ok, for A i got 1/3
(Bu+ C )is ((1/3)u + 2/3)

Answer 12

integrating the first half i get (1/3)ln|u| = (1/3)*ln|e^x|

Answer 13

I'm not getting that for C...
Let me recheck.

Answer 14

oh and I got A=-1/3 but let me recheck

Answer 15

oh, i took the negative out.. if that changes anything/

Answer 16

So you are going to multiply by the negative later.

Answer 17

yeah.. ok so A = 1/3
B = -1/3
C= -2/3

Answer 18

yeah.. it's less confusing for me to take the negative out and multiply by it at the end

Answer 19

So you did this then?
\[u=A(u^2+u+1)+(Bu+C)(u-1)\]
\[u=A(u^2+u+1)+(Bu^2-Bu+Cu-C)\]
\[u=(A+B)u^2+(A-B+C)u+(A-C)\]

Answer 20

\[=>A+B=0 \text{ and } A-B+C=1 \text{ and } A-C=0\]

Answer 21

I'm just decomposing the fraction.
We can come back to the integral later.

Answer 22

I want to see what the integral looks like without substitution at first because she said she didn't want substitution.

Answer 23

We can make substitution later if we need to.

Answer 24

yeah, when solving for A/B/C don't you set it equal to the numerator? so if it were u.. that would mess things up

Answer 25

We are looking at the fraction part right now.

Answer 26

ok anyways...i'm trying to do the problem without substitution because we were told to do without substitution correct?

Answer 27

we can use substitution, i just wrote that after the 'u' substitution (finding a,b and c) I got stuck.. sorry if i wasnt clear!!

Answer 28

and that whole thing is suppose to be negative
not just the first integral

Answer 29

I see what you did there.. ok so when I convert u back to x, the only difference is c which is 2/3

Answer 30

right, i see

Answer 31

now do you want to do sub first then the partial
or the partial then do the sub?

Answer 32

Because I did the partial first because I thought you didn't want to do the sub lol

Answer 33

sub then partial.. so i can follow easier

Answer 34

oh no that x is e^x

Answer 35

\[-\frac{1}{3}( \int\limits\limits_{}^{}\frac{1}{e^x-1} dx +\int\limits\limits_{}^{}\frac{-e^x+1}{e^{2x}+e^x+1}dx) \]

Answer 36

ok so if we look back at integral and not just the fraction
we could do a sub and let u=e^x and so du=e^x dx which is what loser was talking about

Answer 37

\[\int\limits_{}^{}\frac{-e^x}{e^{3x}-1} dx\]\[\int\limits_{}^{}\frac{- du}{u^3-1} =-\int\limits_{}^{}\frac{du}{(u-1)(u^2+u+1)} \]

Answer 38

yeah.. ok so A = 1/3
B = -1/3
C= -2/3
your values in this case are correct

Answer 39

first integral: (1/3)(1/(u-1))

Answer 40

i just quoted what yu said earlier.

Answer 41

awesome!

Answer 42

\[-(\int\limits_{}^{}\frac{1}{3(u-1)} du+\int\limits_{}^{}\frac{\frac{-1}{3}u-\frac{2}{3}}{u^2+u+1} du)\]

Answer 43

your first integral there should be the easier one to deal with...do you need help on the first one?

Answer 44

\[-[-\frac{1}{3}\int\limits_{}^{}\frac{u}{u^2+u+1} du]-[-\frac{2}{3}\int\limits_{}^{}\frac{1}{u^2+u+1} du]\]
I distribute that negative sign we have ...

so on both of thse it looks like you have to do a trig sub

Answer 45

no i got that one - it's -(1/3)*ln|e^x|

Answer 46

wait, how did you change the first integral?

Answer 47

what do you mean?
i didn't rewrite it. i assumed you didn't need help with it.
but shouldn't it be -1/3*ln|u-1|=-1/3*ln|e^x-1|

Answer 48

oh nevermind, ok i and yeah it is u-1, i forgot to add that.. ok so how do you tackle the second one? do you have to do a second partial fraction?

Answer 49

i mentioned trig sub above.

Answer 50

can't factor that is when i hop on the trig sub bandwagon

Answer 51

so you are going to complete the square and find the trig sub.. this problem is ridiculous!

Answer 52

ok, let me try that first

Answer 53

lol yep

Answer 54

\[u^2+u+1=(u^2+u+(\frac{1}{2})^2)+1-(\frac{1}{2})^2=(u+\frac{1}{2})^2+\frac{3}{4}\]

Answer 55

ok, that was killing me.. i was thinking about (u+1)^2 - u.. which wouldn't work.. but this makes sense.. and works

Answer 56

lol
if we had \[ax^2+bx+c\]
on bottom
you would group those parts that the variable part together
\[(ax^2+bx)+c\]
then factor out whatever is in front of the x^2 from both the ax^2 and b term like:
\[a(x^2+\frac{b}{a}x)+c\]
now the completing the square part
if we add something in then we also need to take it out
\[a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)+c-a(\frac{b}{2a})^2 \]
so now we can write
\[a(x+\frac{b}{2a})^2+c-a \frac{b^2}{4a^2}\]
simplify
\[a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}\]

Answer 57

that will be useful.. im going to write this down.. in the meantime.. i worked on the rest and substituted (u+1/2) for h, to make it clearer
so i have 1/(h^2 + 3/4)

Answer 58

next to both integrals

Answer 59

\[(u+1)^2+\frac{3}{4} =\frac{3}{4} ( \frac{4}{3} (u+1)^2+1)=\frac{3}{4}((\frac{2[u+1]}{\sqrt{3}})^2+1)\]
now sub that thing inside the square

Answer 60

by the way I'm just writing the bottom of these integrals.

Answer 61

recall \[\tan^2(\theta)+1=\sec^2(\theta) \]

Answer 62

so tan(theta) is the trig sub we want

Answer 63

ok i'm lost here how did you get (u+1)^2+3/4=(3/4((4/3)(u+1)^2+1)

Answer 64

you took 3/4 out to make the constant 1

Answer 65

I factored out 3/4

Answer 66

but why does 1/2 change in the parenthesis

Answer 67

i wanted it 1 so i could have in the form \[\tan^2(\theta)+1=\sec^2(\theta)\]

Answer 68

\[(u+1)^2+\frac{3}{4} \]
I want something^2+1
\[\frac{4}{3} \cdot \frac{3}{4}(u+1)^2+\frac{3}{4}\]
so i factored 3/4 out
\[\frac{3}{4} \cdot (\frac{4}{3}(u+1)^2+1)\]
that thingy inside the parenthesis is not something^2+1
\[\frac{3}{4} \cdot ([\frac{\sqrt{4}}{\sqrt{3}}(u+1)]^2+1)\]

Answer 69

\[\frac{3}{4} ([\frac{2}{\sqrt{3}}(u+1)]^2+1)\]

Answer 70

you can write it like this:
\[\frac{3}{4}([\frac{2(u+1)}{\sqrt{3}}]^2+1)\]
so it is cleary to you would you want to use as the opposite and the adjacent of some right triangle .

Answer 71

you start off with (u+1/2)^2+(3/4) in the denominator..
but then i see (u+1)^2 +(3/4) which is really confusing me so the rest I can't even try to comprehend until that makes sense lol

Answer 72

yes the 1 is suppose to be 1/2
it gets hard writing in latex sometimes
is that your only question

Answer 73

\[\frac{3}{4}([\frac{2(u+\frac{1}{2})}{\sqrt{3}}]^2+1)\]
if you notice we lost that 1/2 a long time ago
i have been writing 1 for the last 30 minutes on this computer thingy.

Answer 74

yeah, that's why I was incredibly confused.. now it makes sense but i want to run by it again sorry

Answer 75

so \[\frac{ 3 }{ 4 } (\tan^{-1}(\frac{ 2h }{ \sqrt{3} }))\]

Answer 76

\[\text{ let } \tan(\theta)=\frac{2(u+\frac{1}{2})}{\sqrt{3}}\]
\[\sqrt{3} \tan(\theta)=2u+1\]
\[u=\frac{\sqrt{3} \tan(\theta)-1}{2}\]
\[\sec^2(\theta) d \theta=\frac{2}{\sqrt{3}} du\]