What is the integral of 8^{-x} ?

The integral of 8^{-x} is -(8^{-x})/(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 8^{-x}

\int 8^{- x} d x

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

Solution steps

\int 8^{- x} d x

Apply u-substitution

: \int - 8^{u} d u

= \int - 8^{u} d u

Take the constant out:

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

= - \int 8^{u} d u

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

= - \frac{8 ^{u}}{ln ( 8 )}

Substitute back

u = - x

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

Simplify

ln (8) : 3 ln (2)

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{1}{sec ^{2} ( 4 x )})

f (x) = ln (6) - 3 x^{5}

\frac{x ^{2}}{x ^{2} + 1}

x \to \infty lim (52.2 + (41.9 \cdot e^{(- 0.055 \cdot x)}))

\frac{d}{d x} (ln (x^{2} + y))