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

The integral of e^{1/(4x)} is e^{1/(4x)}x-1/4 \Ei(1/(4x))+C

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

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

e^{\frac{1}{4 x}} x - \frac{1}{4} Ei (\frac{1}{4 x}) + C

Solution steps

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

Apply u-substitution

: \int e^{\frac{1}{u}} \frac{1}{4} d u

= \int e^{\frac{1}{u}} \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int e^{\frac{1}{u}} d u

Apply u-substitution

: \int - \frac{e ^{v}}{v ^{2}} d v

= \frac{1}{4} \cdot \int - \frac{e ^{v}}{v ^{2}} d v

Take the constant out:

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

= \frac{1}{4} (- \int \frac{e ^{v}}{v ^{2}} d v)

Apply Integration By Parts

: - \frac{e ^{v}}{v} - \int - \frac{e ^{v}}{v} d v

= \frac{1}{4} (- (- \frac{e ^{v}}{v} - \int - \frac{e ^{v}}{v} d v))

Take the constant out:

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

= \frac{1}{4} (- (- \frac{e ^{v}}{v} - (- \int \frac{e ^{v}}{v} d v)))

Use the nonelementary integral:

\int \frac{e ^{x}}{x} d x = Ei (x)

= \frac{1}{4} (- (- \frac{e ^{v}}{v} - (- Ei (v))))

Substitute back

= \frac{1}{4} (- (- \frac{e ^{\frac{1}{4 x}}}{\frac{1}{4 x}} - (- Ei (\frac{1}{4 x}))))

Simplify

\frac{1}{4} (- (- \frac{e ^{\frac{1}{4 x}}}{\frac{1}{4 x}} - (- Ei (\frac{1}{4 x})))) : e^{\frac{1}{4 x}} x - \frac{1}{4} Ei (\frac{1}{4 x})

= e^{\frac{1}{4 x}} x - \frac{1}{4} Ei (\frac{1}{4 x})

Add a constant to the solution

= e^{\frac{1}{4 x}} x - \frac{1}{4} Ei (\frac{1}{4 x}) + C

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