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

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

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

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

\frac{1}{e ^{y}} ln ∣ 1 + e^{x} ∣ + C

Solution steps

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

Apply u-substitution

= \int \frac{1}{e ^{y} u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{e ^{y}} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{e ^{y}} ln ∣ u ∣

Substitute back

u = 1 + e^{x}

= \frac{1}{e ^{y}} ln ∣ 1 + e^{x} ∣

Add a constant to the solution

= \frac{1}{e ^{y}} ln ∣ 1 + e^{x} ∣ + C

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