What is the integral of e^{2^x} ?

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

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integral of e^{2^x}

\int e^{2^{x}} d x

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

Solution steps

\int e^{2^{x}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the nonelementary integral:

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

= \frac{1}{ln ( 2 )} Ei (u)

Substitute back

u = 2^{x}

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

Add a constant to the solution

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

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