# Integration problem (partial fractions)

#### lha08

1. Homework Statement
Evaluate the integral: integral (-17e^x-36)/(e^(2x)+5e^x+6 dx

2. Homework Equations
partial fractions

3. The Attempt at a Solution
Basically, what i did was factored the bottom into (e^x+2) and (e^x+3) because when i expand that, it equals the bottom. From there, i set A/(e^x+2) + B/(e^x+3) and found that A is -15 and B=-2. As a result, the final answer i got was -15lnabs(e^x+3)-2lnabs(e^x+2)...However it's wrong. Apparently the answer is supposed to be -6x+log(2+e^x)+5log(3+e^x). Help Please!!

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#### Dick

Homework Helper
You should start by doing a change of variables to convert your function to a ratio of polynomials before you do the partial fractions. Like u=e^x. You could also continue with your correct decomposition, but the integral of 1/(e^2+2) is NOT log(e^2+2). You need to do the another u substitution to work that out.

#### lha08

You should start by doing a change of variables to convert your function to a ratio of polynomials before you do the partial fractions. Like u=e^x. You could also continue with your correct decomposition, but the integral of 1/(e^2+2) is NOT log(e^2+2). You need to do the another u substitution to work that out.
When you say change of variables to convert the function to a ratio of polynomials, do you mean that i have to replace all the e^x into a u using u-sub? I'n not that clear on that bit. Also, isn't the integral of 1/(e^2+2) going to involve an arctan since i'm replacing the e^2?

#### HallsofIvy

Homework Helper
Yes. replace every "e^x" with u and, since if u= e^x, du= e^x dx= udx so dx= du/u, replace dx by du/u.

#### Dick

Homework Helper
Sorry, I meant 1/(e^x+2) NOT 1/(e^2+2). But yes, e^x=u and e^(2x)=u^2. du=e^x*dx.

#### Dick

Homework Helper
When you say change of variables to convert the function to a ratio of polynomials, do you mean that i have to replace all the e^x into a u using u-sub? I'n not that clear on that bit. Also, isn't the integral of 1/(e^2+2) going to involve an arctan since i'm replacing the e^2?
Sorry, I meant 1/(e^x+2) NOT 1/(e^2+2). But yes, e^x=u and e^(2x)=u^2. du=e^x*dx.

#### lha08

Sorry, I meant 1/(e^x+2) NOT 1/(e^2+2). But yes, e^x=u and e^(2x)=u^2. du=e^x*dx.
okay so when i replaced the e^x i got.. (-17u-36)/(u^2+5u+6) du/u...and then i integrated (-17u-36)/(u^2+5u+6) using partial fractions and found that A=-2 and B=-15, which gives me -2/(u+2) -15/(u+3) du/u but then when i integrate this, i got (-2lnabs(e^x+2)-15ln(e^x+3)) / e^x which doesn't necessarily compare to the answer that it's suppose to be: -6x+ln(2+e^x)+5ln(3+e^x)

#### Dick

Homework Helper
A/(u+2)+B/(u+3)+C/u. The C/u part is where your -6x is going to come from. You can't just not integrate the du/u.

#### lha08

A/(u+2)+B/(u+3)+C/u. The C/u part is where your -6x is going to come from. You can't just not integrate the du/u.
I'm almost there! but there's still one tiny problem left... i did what you suggested and i did find the correct values for A, B, and C which are 1,5 and -6 but when i replace these values, i got 1/(u+2) +5/(u+3) -6/u. I know how to integrate the first two parts ln(e^x+2)+5ln(e^x+3) but i'm not sure about the -6/u. What I'm thinking is -6ln(u) which is equal to -6ln(e^x). In this case i know that ln(e) is 1 but how does the x come down (suppose to get -6x)?

Homework Helper
ln(a^b)=b*ln(a).

#### lha08

Thanks a lot to everyone who was patient enough to help me especially Dick and HallsofIvy!!

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