# Thread: integration problem (partial fractions)

1. ## integration problem (partial fractions)

i tried this and really have no idea how to approach this particular question:
apparently, i have to first substitute e^x=u and therefore du=e^xdx which gives du/u=dx but when i replace these values, i find that A/(u+2) + B/(u+3)du/u...From there, i find that A=-2 and B=-15 but when i substitute these values, i find (-2ln(e^x+2)-15ln(e^x+3))/e^x. The final answer is supposed to be: -6x+ln(2+e^x)+5ln(3+e^x). Where am i going wrong?
Thanks.

2. You can break this into two separate integrals.
The first one is easy...
$-17\int{e^xdx\over e^{2x}+5e^x+6}$.
Let $u=e^x$ and you have $-17\int{du\over u^2+5u+6}$.
Note that $u^2+5u+6=(u+3)(u+2)$ and that's done.
I can do the secon one, but there must be an easier way.

$-36\int{dx\over e^{2x}+5e^x+6}$ can be done by first multiplying to and bottom by $e^{-x}$
and then letting $u=e^x$ again.
But there should be a better way, something to do with the 17 and 36.

3. consider, for a moment, that you simply did partial fraction decomposition on the integrand, which leaves you needing to solve

$\int \frac{-2}{e^x+2} \, dx + \int \frac{-15}{e^x + 3} \,dx$

Now consider

$\int \frac{1}{e^x+1} \,dx$

$= \int \frac{1 + e^x - e^x}{1+e^x} \,dx$

$= \int 1 - \frac{e^x}{1+e^x} \,dx$

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

Is this hint enough to get you going now?

4. Originally Posted by Stonehambey
consider, for a moment, that you simply did partial fraction decomposition on the integrand, which leaves you needing to solve

$\int \frac{-2}{e^x+2} \, dx + \int \frac{-15}{e^x + 3} \,dx$

Now consider

$\int \frac{1}{e^x+1} \,dx$

$= \int \frac{1 + e^x - e^x}{1+e^x} \,dx$

$= \int 1 - \frac{e^x}{1+e^x} \,dx$

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

Is this hint enough to get you going now?

I like the trick where you multiply the numerator and denominator of

$\int \frac{dx}{e^x+1}$ by $e^{-x}$ giving you $\int \frac{e^{-x}dx}{1+e^{-x}}$ .

Then let $u=1+e^{-x}$.