Integral of e^x: e^x - ln(e^x + 1) + C

[Bushy]

I think this checks out...

\(\displaystyle \int \frac{e^{2x}+e^x-1}{e^x+1}~dx\)

\(\displaystyle \int e^x ~dx- \int \frac{1}{e^x+1}~dx\)

\(\displaystyle e^x-\ln(e^x+1)+C\)

 

[MarkFL]

It checks out iff:

\(\displaystyle \frac{d}{dx}\left(e^x-\ln\left(e^x+1\right)+C\right)=\frac{e^{2x}+e^x-1}{e^x+1}\)

Is this true?

 

[Bushy]

I think it does

 

[MarkFL]

I think it does

I get:

\(\displaystyle \frac{d}{dx}\left(e^x-\ln\left(e^x+1\right)+C\right)=e^x-\frac{e^x}{e^x+1}=\frac{e^{2x}}{e^x+1}\ne\frac{e^{2x}+e^x-1}{e^x+1}\)

If I were going to find the given anti-derivative, I would consider writing the integrand as:

\(\displaystyle \frac{e^{2x}+e^x-1}{e^x+1}=\frac{e^{2x}-1}{e^x+1}+\frac{e^x}{e^x+1}=e^x-1+\frac{e^x}{e^x+1}\)

Now each term can easily be integrated. :)