What is the integral of sqrt(10^{9x)} ?

The integral of sqrt(10^{9x)} is (5^{(9x)/2}*2^{1/2 (2+9x)})/(9ln(10))+C

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integral of sqrt(10^{9x)}

\int 1 0^{9 x} d x

\frac{5 ^{\frac{9 x}{2}} \cdot 2 ^{\frac{1}{2} (2 + 9 x)}}{9 ln ( 10 )} + C

Solution steps

\int 1 0^{9 x} d x

Apply radical rule:

n a^{m} = a^{\frac{m}{n}},

assuming

a \geq 0

= \int 1 0^{\frac{9 x}{2}} d x

Apply u-substitution

: \int 1 0^{u} \cdot \frac{2}{9} d u

= \int 1 0^{u} \cdot \frac{2}{9} d u

Take the constant out:

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

= \frac{2}{9} \cdot \int 1 0^{u} d u

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

= \frac{2}{9} \cdot \frac{1 0 ^{u}}{ln ( 10 )}

Substitute back

u = \frac{9 x}{2}

= \frac{2}{9} \cdot \frac{1 0 ^{\frac{9 x}{2}}}{ln ( 10 )}

Simplify

\frac{2}{9} \cdot \frac{1 0 ^{\frac{9 x}{2}}}{ln ( 10 )} : \frac{5 ^{\frac{9 x}{2}} \cdot 2 ^{\frac{1}{2} (2 + 9 x)}}{9 ln ( 10 )}

= \frac{5 ^{\frac{9 x}{2}} \cdot 2 ^{\frac{1}{2} (2 + 9 x)}}{9 ln ( 10 )}

Add a constant to the solution

= \frac{5 ^{\frac{9 x}{2}} \cdot 2 ^{\frac{1}{2} (2 + 9 x)}}{9 ln ( 10 )} + C

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What is the integral of sqrt(10^{9x)} ?
The integral of sqrt(10^{9x)} is (5^{(9x)/2}*2^{1/2 (2+9x)})/(9ln(10))+C