What is the integral of 2^{sqrt(x)} ?

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

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integral of 2^{sqrt(x)}

\int 2^{x} d x

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

Solution steps

\int 2^{x} d x

Apply u-substitution

: \int 2^{u + 1} u d u

= \int 2^{u + 1} u d u

Apply Integration By Parts

: \frac{2 ^{u + 1} u}{ln ( 2 )} - \int \frac{2 ^{u + 1}}{ln ( 2 )} d u

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

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

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

Substitute back

u = x

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

Add a constant to the solution

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

cos (x) - sin (x)

\frac{d}{d x} ((x^{2} - 1)^{\frac{1}{2}})

\int 3 (1 + x^{2})^{- 1} d x

\int e^{z + e^{z}} d z

\frac{d}{d x} (e^{15 x})

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