What is the integral of 2^x2^{2^x} ?

The integral of 2^x2^{2^x} is 1/(ln^2(2))*2^{2^x}+C

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integral of 2^x2^{2^x}

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

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

Solution steps

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

Apply u-substitution

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

\int a^{x} d x = \frac{a ^{x}}{ln a}

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

Substitute back

u = 2^{x}

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

Simplify

\frac{1}{ln ( 2 )} \cdot \frac{2 ^{2^{x}}}{ln ( 2 )} : \frac{1}{ln ^{2} ( 2 )} \cdot 2^{2^{x}}

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

Add a constant to the solution

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

What is the integral of 2^x2^{2^x} ?
The integral of 2^x2^{2^x} is 1/(ln^2(2))*2^{2^x}+C