Find int [ (e^x) / (e^2x - 9) ] dx

[mooshupork34]

Find the integral: int [ (e^x) / (e^(2x) - 9) ] dx.

First, I figured that e^(2x) is the same as (e^x)^2. Therefore, I changed the integrand to:

. . .(e^x) * ((e^x)^2 - 9)^(-1)

Using u-substitution, I then set u equal to e^x. Because dx is then equal to 1/(e^x) du, I multiplied that by the rest of the integral: e^x multiplied by 1 over itself cancels out, and I was left with:

. . .int [ ((u^2) - 9)^(-1) ] du

This is where I got stuck. Advice?

Thank you!

 

[soroban]

Re: integrating

Hello, mooshupork34!

i was asked to find: \(\displaystyle \L\:\int\frac{e^x}{e^{2x}\,-\,9}\,dx\)

So this is what i have so far.
. . i figured that $\displaystyle e^{2x}$ is the same as $\displaystyle (e^x)^2$

i changed the function to: \(\displaystyle \L\:\frac{e^x\.dx}{(e^x)^2\,-\,9\)

i then set $\displaystyle u\,=\,e^x$

because dx is equal to $\displaystyle \frac{du}{e^x}$, i multiplied that by the rest of the integral.
$\displaystyle e^x$ multiplied by $\displaystyle \frac{1}{e^x}$ cancels out.

And i was left with: \(\displaystyle \L\:\int\frac{du}{u^2\,-\,9}\)

This is where i got stuck.

There are at least three ways to integrate this . . .

[1] Let $\displaystyle u\:=\;3\sec\theta$ and use Trig Substitution.

[2] We have: \(\displaystyle \L\:\frac{1}{(u\,-\,3)(u\,+\,3)}\) . . . use Partial Fractions.

[3] Simply know the formula for this integral: \(\displaystyle \L\:\int \frac{du}{u^2\,-\,a^2} \;=\;\frac{1}{2a}\ln\left|\frac{u\,-\,a}{u\,+\,a}\right| \,+\,C\)