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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\int x ln (d) x d x

\int sec^{2} (5 x - 1) d x

\int \frac{- 2 x ^{2}}{x ^{2} - 2 x - 48} d x

\int 3 x + cos (x) d x

\int (3 x + 1) sin (x) d x

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