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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Apply u-substitution

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

Take the partial fraction of

\frac{1}{v ( v ^{2} + v )} : - \frac{1}{v} + \frac{1}{v ^{2}} + \frac{1}{v + 1}

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

\int \frac{1}{v} d v = ln ∣ v ∣

\int \frac{1}{v ^{2}} d v = - \frac{1}{v}

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

= 2 (- ln ∣ v ∣ - \frac{1}{v} + ln ∣ v + 1 ∣)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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