What is the Integral of (1+e^x)/(1-e^x)?

[nameVoid]

<br /> \int \frac{1+e^x}{1-e^x}dx<br />
<br /> \int \frac{dx}{1-e^x} +\int \frac{e^x}{1-e^x}dx<br />
<br /> u=e^x <br />
<br /> lnu=x<br />
<br /> \frac{du}{u}=dx<br />
<br /> \int \frac{du}{u(1-u)}+\int \frac{du}{1-u}<br />
<br /> \int \frac{A}{u}+\frac{B}{1-u}du -ln|1-u|+C<br />
<br /> ln|e^x|+ln|1-e^x|-ln|1-e^x|+C<br />
<br /> x+C<br />

 

[Mark44]

<br /> \int \frac{1+e^x}{1-e^x}dx<br />
<br /> \int \frac{dx}{1-e^x} +\int \frac{e^x}{1-e^x}dx<br />
<br /> u=e^x <br />
<br /> lnu=x<br />
<br /> \frac{du}{u}=dx<br />
<br /> \int \frac{du}{u(1-u)}+\int \frac{du}{1-u}<br />
<br /> \int \frac{A}{u}+\frac{B}{1-u}du -ln|1-u|+C<br />
<br /> ln|e^x|+ln|1-e^x|-ln|1-e^x|+C<br />
<br /> x+C<br />

Your antiderivative is obviously incorrect, since d/dx(x + C) = 1. Your antiderivative would have been correct if its derivative was (1 + e^x)/(1 - e^x).

It might be easier not to split into two integrals, but using the same substitution. If you do that, you'll get something you can use partial decomposition on.

 

[nameVoid]

<br /> x-2ln|1-e^x|+C<br />