# Misc integral

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

## The Attempt at a Solution

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Mark44
Mentor
$$\int \frac{1+e^x}{1-e^x}dx$$
$$\int \frac{dx}{1-e^x} +\int \frac{e^x}{1-e^x}dx$$
$$u=e^x$$
$$lnu=x$$
$$\frac{du}{u}=dx$$
$$\int \frac{du}{u(1-u)}+\int \frac{du}{1-u}$$
$$\int \frac{A}{u}+\frac{B}{1-u}du -ln|1-u|+C$$
$$ln|e^x|+ln|1-e^x|-ln|1-e^x|+C$$
$$x+C$$
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.

$$x-2ln|1-e^x|+C$$