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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ( 1 + u )} d u

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

Apply u-substitution

: \int \frac{2}{v ( v + 1 )} d v

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

Take the constant out:

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

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

Take the partial fraction of

\frac{1}{v ( v + 1 )} : \frac{1}{v} - \frac{1}{v + 1}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{v} d v = ln ∣ v ∣

\int \frac{1}{v + 1} d v = ln ∣ v + 1 ∣

= 2 (ln ∣ v ∣ - ln ∣ v + 1 ∣)

Substitute back

= 2 (ln e^{x} - ln e^{x} + 1)

Simplify

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

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

Add a constant to the solution

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

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