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

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

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integral of 1/(e^x+e^x)

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

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

Solution steps

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

Apply Integration By Parts

: \frac{x}{2 e ^{x}} - \int - \frac{x}{2 e ^{x}} d x

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

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

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

Simplify

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

Add a constant to the solution

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

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