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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

= \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

\int \frac{3 x ^{2} - 5 x}{x} d x

e^{- 4 x}

\int (\frac{x ^{2} + 1}{x})^{2} d x

\int (10 x - 10) sin^{2} (5 x^{2} - 10 x + 1) d x

\int \frac{1}{1 - 6 x - 3 x ^{2}} d x

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