What is the integral of (6^{sqrt(x)})/(sqrt(x)) ?

The integral of (6^{sqrt(x)})/(sqrt(x)) is (2*6^{sqrt(x)})/(ln(6))+C

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integral of (6^{sqrt(x)})/(sqrt(x))

\int \frac{6 ^{x}}{x} d x

\frac{2 \cdot 6 ^{x}}{ln ( 6 )} + C

Solution steps

\int \frac{6 ^{x}}{x} d x

Apply u-substitution

: \int 3^{u} \cdot 2^{u + 1} d u

= \int 3^{u} \cdot 2^{u + 1} d u

Simplify

2^{u + 1} : 2^{u} \cdot 2

= \int 3^{u} \cdot 2^{u} \cdot 2 d u

Take the constant out:

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

= 2 \cdot \int 3^{u} \cdot 2^{u} d u

Simplify

3^{u} \cdot 2^{u} : 6^{u}

= 2 \cdot \int 6^{u} d u

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

= 2 \cdot \frac{6 ^{u}}{ln ( 6 )}

Substitute back

u = x

= 2 \cdot \frac{6 ^{x}}{ln ( 6 )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 \cdot 6 ^{x}}{ln ( 6 )}

Add a constant to the solution

= \frac{2 \cdot 6 ^{x}}{ln ( 6 )} + C

\int \frac{3 x ^{2}}{x ^{2} - x - 12} d x

\frac{\partial}{\partial z} (tan (2 + 2 x^{2} y^{4} z^{2}))

\frac{d}{d x} (sin (x) \cdot cos (y))

n = 1 \sum \infty - \frac{2}{n}

n = 1 \sum \infty \frac{1}{n} (\frac{x}{5})^{n}

What is the integral of (6^{sqrt(x)})/(sqrt(x)) ?
The integral of (6^{sqrt(x)})/(sqrt(x)) is (2*6^{sqrt(x)})/(ln(6))+C