What is the integral of (3^x)/(sqrt(1-9^x)) ?

The integral of (3^x)/(sqrt(1-9^x)) is (arcsin(3^x))/(ln(3))+C

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integral of (3^x)/(sqrt(1-9^x))

\int \frac{3 ^{x}}{1 - 9 ^{x}} d x

\frac{arcsin ( 3 ^{x} )}{ln ( 3 )} + C

Solution steps

\int \frac{3 ^{x}}{1 - 9 ^{x}} d x

Apply u-substitution

: \int - \frac{1}{2 ln ( 3 ) \cdot 3 ^{\frac{l n ( - u + 1 )}{2 l n ( 3 )}} u} d u

= \int - \frac{1}{2 ln ( 3 ) \cdot 3 ^{\frac{l n ( - u + 1 )}{2 l n ( 3 )}} u} d u

Take the constant out:

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

= - \frac{1}{2 ln ( 3 )} \cdot \int \frac{1}{3 ^{\frac{l n ( - u + 1 )}{2 l n ( 3 )}} u} d u

Apply u-substitution

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

= - \frac{1}{2 ln ( 3 )} \cdot \int - \frac{2}{- v ^{2} + 1} d v

Take the constant out:

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

= - \frac{1}{2 ln ( 3 )} (- 2 \cdot \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = arcsin (v)

= - \frac{1}{2 ln ( 3 )} (- 2 arcsin (v))

Substitute back

= - \frac{1}{2 ln ( 3 )} (- 2 arcsin (3^{\frac{l n ( - ( 1 - 9 ^{x} ) + 1 )}{2 l n ( 3 )}}))

Simplify

- \frac{1}{2 ln ( 3 )} (- 2 arcsin (3^{\frac{l n ( - ( 1 - 9 ^{x} ) + 1 )}{2 l n ( 3 )}})) : \frac{arcsin ( 3 ^{x} )}{ln ( 3 )}

= \frac{arcsin ( 3 ^{x} )}{ln ( 3 )}

Add a constant to the solution

= \frac{arcsin ( 3 ^{x} )}{ln ( 3 )} + C

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What is the integral of (3^x)/(sqrt(1-9^x)) ?
The integral of (3^x)/(sqrt(1-9^x)) is (arcsin(3^x))/(ln(3))+C