# Integration by partial fractions help?

• Apr 11th 2014, 11:16 PM
jiel
Integration by partial fractions help?
I'm trying to find the integral of

(-12e^x-20)/(e^(2x)+6e^x+5)

using integration by partial fractions, but I keep getting stuck at

-4 (/int/ (1/2)/(e^x+1) + /int/ (5/2)/(e^x+5))

I'm not sure if I messed up on one of the intermediate steps. If not, how do I proceed from here to finish integrating?

Note: the answer is supposed to be -4x+2ln(1+e^x)+2ln(5+e^x), but I have no idea how to get there.

Any help would be appreciated, thank you!
• Apr 11th 2014, 11:52 PM
romsek
Re: Integration by partial fractions help?
Quote:

Originally Posted by jiel
I'm trying to find the integral of

(-12e^x-20)/(e^(2x)+6e^x+5)

using integration by partial fractions, but I keep getting stuck at

-4 (/int/ (1/2)/(e^x+1) + /int/ (5/2)/(e^x+5))

I'm not sure if I messed up on one of the intermediate steps. If not, how do I proceed from here to finish integrating?

Note: the answer is supposed to be -4x+2ln(1+e^x)+2ln(5+e^x), but I have no idea how to get there.

Any help would be appreciated, thank you!

$f(x)=\dfrac{-12 e^x - 20}{e^{2x}+6e^x+5}$

let $u=e^x$

$f(u)=\dfrac{-12u-20}{u^2+6u+5}$

$f(u)=\dfrac{-12u-20}{(u+5)(u+1)}$

$f(u) = \dfrac A {u+5} + \dfrac B {u+1}$

$A(u+1) + B(u+5) = -12u - 20$

You should be able to finish this from here. Just substitute back $e^x$ for $u$ when you are finished.
• Apr 12th 2014, 02:09 PM
Soroban
Re: Integration by partial fractions help?
Hello, jiel!

Quote:

I'm trying to find the integral of: $\frac{-12e^x-20}{e^{2x}+6e^x+5}$ with partial fractions

but I keep getting stuck at: . $-4\int\left(\frac{\frac{1}{2}}{e^x+1} +\frac{\frac{5}{2}}{e^x+5}\right)dx$

You have: . $-2\int\left(\frac{1}{e^x+1} + \frac{5}{e^x+5}\right)dx$

Watch this!

Consider: . $I \;=\;\int\frac{dx}{e^x+1}$

Multiply by $\frac{e^{-x}}{e^{-x}}\!:\;\;I \;=\;\int\frac{e^{-x}}{e^{-x}}\cdot\frac{dx}{e^x+1} \;=\;\int\frac{e^{-x}dx}{1 + e^{-x}}$

Let $u \,=\, 1+e^{-x} \quad\Rightarrow\quad du \,=\,-e^{-x}dx \quad\Rightarrow\quad e^{-x}dx \,=\,-du$

Substitute: . $I \;=\;\int\frac{-du}{1+u} \;=\;-\int\frac{du}{1+u} \;=\;-\ln|1+u| + C$

Back-substitute: . $I \;=\;-\ln\left(1 + e^{-x}\right) + C$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This can be simplified beyond all recognition . . .

$-\ln(1 + e^{-x}) + C$

. . . $=\;-\ln\left(1 + \frac{1}{e^x}\right) + C$

. . . $=\;-\ln\left(\frac{e^x+1}{e^x}\right) + C$

. . . $=\;\ln\left(\frac{e^x+1}{e^x}\right)^{-1} + C$

. . . $=\; \ln\left(\frac{e^x}{e^x+1}\right)+C$

. . . $=\;\ln(e^x) - \ln(e^x+1) + C$

. . . $=\;x - \ln(e^x+1) + C$