What is the integral of 2^{1/x} ?

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

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integral of 2^{1/x}

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

2^{\frac{1}{x}} x - ln (2) Ei (ln (2^{\frac{1}{x}})) + C

Solution steps

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

Apply u-substitution

: \int - \frac{ln ( 2 )}{ln ^{2} ( u )} d u

= \int - \frac{ln ( 2 )}{ln ^{2} ( u )} d u

Take the constant out:

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

= - ln (2) \cdot \int \frac{1}{ln ^{2} ( u )} d u

Apply u-substitution

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

= - ln (2) \cdot \int \frac{e ^{v}}{v ^{2}} d v

Apply Integration By Parts

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

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

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

= - ln (2) (- \frac{e ^{v}}{v} - (- Ei (v)))

Substitute back

= - ln (2) - \frac{e ^{l n (2^{\frac{1}{x}})}}{ln ( 2 ^{\frac{1}{x}} )} - (- Ei (ln (2^{\frac{1}{x}})))

Simplify

- ln (2) - \frac{e ^{l n (2^{\frac{1}{x}})}}{ln ( 2 ^{\frac{1}{x}} )} - (- Ei (ln (2^{\frac{1}{x}}))) : 2^{\frac{1}{x}} x - ln (2) Ei (ln (2^{\frac{1}{x}}))

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

Add a constant to the solution

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

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