What is the integral of (4^{sqrt(x)})/(sqrt(x)) ?

The integral of (4^{sqrt(x)})/(sqrt(x)) is (4^{sqrt(x)})/(ln(2))+C

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integral of (4^{sqrt(x)})/(sqrt(x))

\int \frac{4 ^{x}}{x} d x

\frac{4 ^{x}}{ln ( 2 )} + C

Solution steps

\int \frac{4 ^{x}}{x} d x

Apply u-substitution

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

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

Apply u-substitution

: \int 2^{v - 1} d v

= \int 2^{v - 1} d v

Simplify

2^{v - 1} : 2^{v} \cdot 2^{- 1}

= \int 2^{v} \cdot 2^{- 1} d v

Take the constant out:

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

= 2^{- 1} \cdot \int 2^{v} d v

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{2} \cdot \int 2^{v} d v

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

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

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{2 ^{2 x + 1}}{ln ( 2 )} : \frac{4 ^{x}}{ln ( 2 )}

= \frac{4 ^{x}}{ln ( 2 )}

Add a constant to the solution

= \frac{4 ^{x}}{ln ( 2 )} + C

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\frac{\partial}{\partial x} (3 e^{- y} (x^{2} + y^{2}) + 4)

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What is the integral of (4^{sqrt(x)})/(sqrt(x)) ?
The integral of (4^{sqrt(x)})/(sqrt(x)) is (4^{sqrt(x)})/(ln(2))+C