Integration by Substitution..

[hank]

Ok, here's my problem:

\(\displaystyle \int_{ - \ln 3}^{\ln 3} {\frac{{e^x }}{{e^x + 4}}} dx\)

I've tried this problem about five different times, with no luck.

Can anyone help?

I'm setting up my u for the substitution to be the e^x + 4.

I end up with:

\(\displaystyle \ln \frac{{e^7 + 4}}{{e^{\frac{{13}}{3}} + 4}}\)

Which the book tells me is wrong.

 

[soroban]

Hello, Hank!

\(\displaystyle \L\int^{\;\;\;\;\ln3}_{-\ln3}\frac{e^x}{e^x\,+\,4}\,dx\)

If we let \(\displaystyle u\:=\:e^x\,+\,4\), can't you see that the numerator is \(\displaystyle du\) ?


So the integral is: \(\displaystyle \L\:\int\frac{du}{u}\:=\:\ln(u) \,+\,C\)

Back-substitute: \(\displaystyle \:\ln\left(e^x\,+\,4\right)\,\bigg]^{\ln3}_{-\ln3}\)

We have: \(\displaystyle \:\ln\left(e^{^{\ln 3}}\,+\,4\right) \,- \,\ln\left(e^{^{-\ln3}}\,+\,4\right)\;=\;\ln\left(3\,+\,4\right) \,- \,\ln\left(\frac{1}{3}\,+\,4\right)\;\) *

. . \(\displaystyle =\;\ln\left(7\right) \,-\,\ln\left(\frac{13}{3}\right) \;=\;\ln\left(\frac{7}{\frac{13}{3}}\right) \;= \;\L\fbox{\ln\left(\frac{21}{13}\right)}\)

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

*

We had: \(\displaystyle \L\,e^{(-\ln 3)} \;=\;e^{(\ln^{3^{-1}})} \;=\;e^{\ln(\frac{1}{3})}\;=\;\frac{1}{3}\)

 

[hank]

Thanks for the help!

I see now where my problem was.

I was subbing back in wrong.

Thanks!