What is the integral of 9^{1-4x} ?

The integral of 9^{1-4x} is -(3^{-8x+2})/(8ln(3))+C

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integral of 9^{1-4x}

\int 9^{1 - 4 x} d x

- \frac{3 ^{- 8 x + 2}}{8 ln ( 3 )} + C

Solution steps

\int 9^{1 - 4 x} d x

9^{1 - 4 x} = 9^{1} \cdot 9^{- 4 x}

= \int 9^{1} \cdot 9^{- 4 x} d x

Take the constant out:

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

= 9^{1} \cdot \int 9^{- 4 x} d x

Simplify

= 9 \cdot \int 9^{- 4 x} d x

Simplify

9^{- 4 x} : 656 1^{- x}

= 9 \cdot \int 656 1^{- x} d x

Apply u-substitution

: \int - 656 1^{u} d u

= 9 \cdot \int - 656 1^{u} d u

Take the constant out:

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

= 9 (- \int 656 1^{u} d u)

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

= 9 (- \frac{656 1 ^{u}}{ln ( 6561 )})

Substitute back

u = - x

= 9 (- \frac{656 1 ^{- x}}{ln ( 6561 )})

Simplify

9 (- \frac{656 1 ^{- x}}{ln ( 6561 )}) : - \frac{3 ^{- 8 x + 2}}{8 ln ( 3 )}

= - \frac{3 ^{- 8 x + 2}}{8 ln ( 3 )}

Add a constant to the solution

= - \frac{3 ^{- 8 x + 2}}{8 ln ( 3 )} + C

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What is the integral of 9^{1-4x} ?
The integral of 9^{1-4x} is -(3^{-8x+2})/(8ln(3))+C