What is the integral of 1/(e^{x/2)+e^{x/4}} ?

The integral of 1/(e^{x/2)+e^{x/4}} is -x-4/(e^{x/4)}+4ln|e^{x/4}+1|+C

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integral of 1/(e^{x/2)+e^{x/4}}

\int \frac{1}{e ^{\frac{x}{2}} + e ^{\frac{x}{4}}} d x

- x - \frac{4}{e ^{\frac{x}{4}}} + 4 ln e^{\frac{x}{4}} + 1 + C

Solution steps

\int \frac{1}{e ^{\frac{x}{2}} + e ^{\frac{x}{4}}} d x

Apply u-substitution

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

Apply u-substitution

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

Take the constant out:

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

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

Apply u-substitution

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

Take the partial fraction of

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

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

Apply the Sum Rule:

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

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

\int \frac{1}{w} d w = ln ∣ w ∣

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

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

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

Substitute back

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

Simplify

2 \cdot 2 (- ln e^{\frac{\frac{x}{2}}{2}} - \frac{1}{e ^{\frac{\frac{x}{2}}{2}}} + ln e^{\frac{\frac{x}{2}}{2}} + 1) : - x - \frac{4}{e ^{\frac{x}{4}}} + 4 ln e^{\frac{x}{4}} + 1

= - x - \frac{4}{e ^{\frac{x}{4}}} + 4 ln e^{\frac{x}{4}} + 1

Add a constant to the solution

= - x - \frac{4}{e ^{\frac{x}{4}}} + 4 ln e^{\frac{x}{4}} + 1 + C

What is the integral of 1/(e^{x/2)+e^{x/4}} ?
The integral of 1/(e^{x/2)+e^{x/4}} is -x-4/(e^{x/4)}+4ln|e^{x/4}+1|+C