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

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

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

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

\frac{1}{e} arctan (e^{x - 1}) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2} + e ^{2}} d u

= \int \frac{1}{u ^{2} + e ^{2}} d u

u^{2} + e^{2} = e^{2} (\frac{u ^{2}}{e ^{2}} + 1)

= \int \frac{1}{e ^{2} ( \frac{u ^{2}}{e ^{2}} + 1 )} d u

Take the constant out:

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

= \frac{1}{e ^{2}} \cdot \int \frac{1}{\frac{u ^{2}}{e ^{2}} + 1} d u

Apply Integral Substitution

= \frac{1}{e ^{2}} \cdot \int \frac{e}{v ^{2} + 1} d v

Take the constant out:

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

= \frac{1}{e ^{2}} e \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{e ^{2}} e arctan (v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{e} arctan (e^{x - 1}) + C

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