Help with an integral

[wondering]

I need help with where to start
\(\displaystyle \int\frac{5}{3e^x-2}dx\)

\(\displaystyle 5\int\frac{dx}{3e^x-2}\)
I pulled the 5 out, but I am not sure what to do next.

 

[srmichael]

I need help with where to start
\(\displaystyle \int\frac{5}{3e^x-2}dx\)

\(\displaystyle 5\int\frac{dx}{3e^x-2}\)
I pulled the 5 out, but I am not sure what to do next.

Yes, this is a tricky one. I normally don't go "Soroban" and post the entire solution, but I'm making an exception here.

\(\displaystyle 5\int\frac{dx}{3e^x-2}\)

Let \(\displaystyle u=e^x\)
Then \(\displaystyle du=e^xdx\)
Thus \(\displaystyle dx=\frac{du}{e^x}\)

Therefore: \(\displaystyle 5\int\frac{dx}{3e^x-2}=5\int\frac{du}{u(3u-2)}\)

Use Partial Fractions:

\(\displaystyle \frac{1}{u(3u-2)}=\frac{A}{u}+\frac{B}{3u-2}\)
\(\displaystyle 1=A(3u-2)+Bu\)
\(\displaystyle 1=3Au-2A+Bu\)
\(\displaystyle 1=(3A+B)u-2A\)
\(\displaystyle -2A=1\)
\(\displaystyle A=-\frac{1}{2}\)
\(\displaystyle 3A+B=0\)
\(\displaystyle 3(-\frac{1}{2})+B=0\)
\(\displaystyle B=\frac{3}{2}\)

So: \(\displaystyle 5\int\frac{du}{u(3u-2)}=5\int(-\frac{1}{2u}+\frac{3}{2(3u-2)})du\)
\(\displaystyle =-\frac{5}{2}\int\frac{du}{u}+\frac{15}{2}\int\frac{du}{3u-2}\)

Let \(\displaystyle w = 3u-2\)
Then \(\displaystyle dw=3du\)
Thus \(\displaystyle du=\frac{dw}{3}\)

Therfore:\(\displaystyle -\frac{5}{2}\int\frac{du}{u}+\frac{15}{2}\int\frac{du}{3u-2}=-\frac{5}{2}\int\frac{du}{u}+\frac{15}{2}\int\frac{dw}{3w}\)
\(\displaystyle =-\frac{5}{2}\int\frac{du}{u}+\frac{5}{2}\int\frac{dw}{w}\)
\(\displaystyle =-\frac{5}{2}\ln(u)+\frac{5}{2}\ln(w)+C\)
\(\displaystyle =-\frac{5}{2}\ln(u)+\frac{5}{2}\ln(3u-2)+C\)
\(\displaystyle =-\frac{5}{2}\ln(e^x)+\frac{5}{2}\ln(3e^x-2)+C\)
\(\displaystyle =\frac{5}{2}[\ln(3e^x-2)-x]+C\)

 

[DrPhil]

I need help with where to start
\(\displaystyle \int\frac{5}{3e^x-2}dx\)

\(\displaystyle 5\int\frac{dx}{3e^x-2}\)
I pulled the 5 out, but I am not sure what to do next.

I enjoyed reading srmichael's reply, but perhaps I have found an easier way. Multiply numerator and denominator by \(\displaystyle e^{-x}\):

\(\displaystyle \displaystyle \int\frac{5}{3e^x-2}dx = \int\frac{5e^{-x}}{3-2e^{-x}}dx\)

Now let \(\displaystyle \displaystyle u = e^{-x}\),......\(\displaystyle du = - e^{-x}\ dx \)

.......\(\displaystyle \displaystyle \int \dfrac{-5\ du}{3 - 2\ u}\)

The integral is a logarithm. Since there is a minus sign in the argument, better make it absolute value. Should wind up with same result as srmichael - perhaps arranged differently because f using e^{-x} instead of e^x.

 

[wondering]

Thank you both for the replies. I will have to review both of your suggestions when I get home from work tonight to see if I understand what you did. I may have to ask a few more questions tonight. Thanks again.

 

[HallsofIvy]

My first thought would have been to substitute for the entire denominator: let \(\displaystyle u= 3e^x -2\). Then \(\displaystyle du= 3e^x dx\) so that \(\displaystyle dx= \frac{du}{3e^x}= \frac{du}{u+ 2}\). The integral becomes
\(\displaystyle \int\frac{du}{u(u+2)}\) and now you can use partial fractions.