Question

Partial fraction help

Answer 1

\[\int\limits \frac{ e^(2x) }{ e^(2x)-e^x -42}dx\]

Answer 2

for the \[e^(2x)\] the 2x is together sorry

Answer 3

Have you been able to factor the denominator?

Answer 4

would they just be e^2x and e^x after you bring out the (1/-42)?

Answer 5

If you wanna use partial fractions, you'll have to use long division first to make the power less on the top.

Answer 6

or do we factor out the whole thing

Answer 7

its hard

Answer 8

bye

Answer 9

Well, the denominator is a quadratic, its just quadratic in e^x instead of x. So you can factor it like any quadratic. If it helps, you can use a temporary change of variable to make it look more friendly. We can say let e^x = y. Then we have:
\[\int\limits_{}^{}\frac{ y^{2} }{ y^{2}-y-42 }dy\]. Would you be able to factor a denominator that looked like that?

Answer 10

(e^x-7)(e^x+6)

Answer 11

Correct. Now e^x isnt linear, but we can still do this partial fractions as if these were linear factors. Do you know how to set up a partial fraction problem with linear factors?

Answer 12

\[e^(2x)= A(e^x-7)+B(e^x+6)\]

Answer 13

opps swap that

Answer 14

\[e^(2x)=A(e^x+6)+B(e^x-7)\]

Answer 15

I don't know what to plug into x though

Answer 16

Right. So since these are equal, this equation will be true for all x values. Now, we wont be able to plug in a fancy value for the e^x + 6 factor, but we can choose an x value that will work for the e^x - 7 factore. We can make e^x - 7 equal 0 if we let x = ln7.

Answer 17

what about trig sub

Answer 18

\[\int\limits_{}^{} \frac{e^{2x}}{(e^x-\frac{1}{2})^2-\frac{169}{4}} dx=\int\limits_{}^{}\frac{e^{2x } dx}{(e^x-\frac{1}{2})^2-(\frac{13}{2})^2} \\ =(\frac{2}{13})^2 \int\limits_{}^{}\frac{e^{2x}}{[\frac{2}{13}(e^x-\frac{1}{2})]^2-1} dx\]

Answer 19

Are you allowed to do this with other methods, shadow? If so, I would probably suggest trig sub as well.

Answer 20

but i guess we are suppose to do this problem by partial fractions alone?

Answer 21

so A=49/13
and B=-36?

Answer 22

\[\int\limits \frac{ e^{2x} }{ e^{2x}-e^x -42}dx = \int\limits1 + \frac{ e^{x}+42 }{ e^{2x}-e^x -42}dx\]

Answer 23

as luigi said earlier, you need to diminish the power of numerator for partial fractions to work

Answer 24

\[
\frac{ y^{2} }{ y^{2}-y-42 } = \frac{ (y^{2}-y-42)+(y+42) }{ y^{2}-y-42 } =
1 + \frac{ y+42 }{ y^{2}-y-42 } = \\
1 + \frac{ y+42 }{(y-7)(y+6) }
\]

Answer 25

um, we can use others as well if its easier :/

Answer 26

Hm, I was never taught that you had to reduce the numerator, lol. Everytime there was a partial fraction problem, I guess the power never needed to be reduced, so I never caught on to needing to have it done. Well, I'm clearly the idiot here, so I'll let others help before I mislead you again, sorry.

Answer 27

\[\int\limits_{}^{}\frac{A(e^x+7)}{e^{2x}-49} dx=\int\limits_{}^{}\frac{Ae^x}{e^{2x}-49} dx+\int\limits_{}^{}\frac{A7}{e^{2x}-49} dx\]
you still need a trig sub

Answer 28

e^x/7=sec(theta)
tan(theta) d theta=dx <--you can show this by manipulating your derivative with your sub

Answer 29

\[
1 + \frac{ y+42 }{(y-7)(y+6) } = 1 + \frac{49}{13(y-7)} - \frac{36}{13(y+6)} \\
\frac{ e^{2x} }{ e^{2x}-e^x -42} = 1 + \frac{49}{13(e^x-7)} - \frac{36}{13(e^x+6)} \\
\]

Answer 30

BTW, shadow... Enclose the exponents within curly braces in LaTex to get proper output. e^{2x}

Answer 31

ok

Answer 32

So putting it in partial fractions as shown above does not make the integration any easier. However, putting it in partial fractions a bit differently may help. We need an \(e^x\) in the numerator with each partial fraction:
\[
\frac{ e^{2x} }{ e^{2x}-e^x -42} = \frac{ e^{2x} }{(e^x-7)(e^x+6)} =
\frac{Ae^x}{e^x-7} + \frac{Be^x}{e^x+6} = \\
\frac{Ae^{2x}+6Ae^x+Be^{2x}-7Be^x}{(e^x-7)(e^x+6)} =
\frac{(A+B)e^{2x}+(6A-7B)e^x}{(e^x-7)(e^x+6)} \\
\frac{ e^{2x} }{(e^x-7)(e^x+6)} = \frac{(A+B)e^{2x}+(6A-7B)e^x}{(e^x-7)(e^x+6)} \\
\text{ } \\
A+B = 1; ~~ 6A-7B = 0. ~~~~7A + 7B = 7; ~~13A = 7; ~~\\
\text{ } \\
A = \frac{7}{13}; ~~ B = 1 - \frac{7}{13} = \frac{6}{13} \\
\text{ } \\
\int \frac{ e^{2x} }{ e^{2x}-e^x -42}dx = \frac{7}{13}\int \frac{e^x}{e^x-7}dx +
\frac{6}{13}\int \frac{e^x}{e^x+6}dx = \\
\text{ } \\
\frac{7}{13}\ln(e^x-7) + \frac{6}{13}\ln(e^x+6) + C

\]