What is the integral of (e^{x-1})/(e^{x+1)} ?

The integral of (e^{x-1})/(e^{x+1)} is 1/(e^2)x+C

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integral of (e^{x-1})/(e^{x+1)}

\int \frac{e ^{x - 1}}{e ^{x + 1}} d x

\frac{1}{e ^{2}} x + C

Solution steps

\int \frac{e ^{x - 1}}{e ^{x + 1}} d x

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

= \int \frac{1}{e ^{2}} d x

Integral of a constant:

\int a d x = a x

= \frac{1}{e ^{2}} x

Add a constant to the solution

= \frac{1}{e ^{2}} x + C

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What is the integral of (e^{x-1})/(e^{x+1)} ?
The integral of (e^{x-1})/(e^{x+1)} is 1/(e^2)x+C