What is the integral of (1/4)^{3x} ?

The integral of (1/4)^{3x} is -(2^{-6x-1})/(3ln(2))+C

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integral of (1/4)^{3x}

\int (\frac{1}{4})^{3 x} d x

- \frac{2 ^{- 6 x - 1}}{3 ln ( 2 )} + C

Solution steps

\int (\frac{1}{4})^{3 x} d x

Apply exponent rule:

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

\frac{1}{4} = 4^{- 1}

= \int (4^{- 1})^{3 x} d x

Simplify

(4^{- 1})^{3 x} : 6 4^{- x}

= \int 6 4^{- x} d x

Apply u-substitution

= \int - 6 4^{u} d u

Take the constant out:

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

= - \int 6 4^{u} d u

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

= - \frac{6 4 ^{u}}{ln ( 64 )}

Substitute back

u = - x

= - \frac{6 4 ^{- x}}{ln ( 64 )}

Simplify

- \frac{6 4 ^{- x}}{ln ( 64 )} : - \frac{2 ^{- 6 x - 1}}{3 ln ( 2 )}

= - \frac{2 ^{- 6 x - 1}}{3 ln ( 2 )}

Add a constant to the solution

= - \frac{2 ^{- 6 x - 1}}{3 ln ( 2 )} + C

What is the integral of (1/4)^{3x} ?
The integral of (1/4)^{3x} is -(2^{-6x-1})/(3ln(2))+C