What is the integral of (sqrt(e^x))/(2(1+e^x)) ?

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

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integral of (sqrt(e^x))/(2(1+e^x))

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

arctan (e^{\frac{x}{2}}) + C

Solution steps

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

Take the constant out:

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

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

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{1 + v ^{2}} d v

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

Use the common integral:

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

= \frac{1}{2} \cdot 2 arctan (v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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