What is the integral of 8/(5^{3x)} ?

The integral of 8/(5^{3x)} is -(8*125^{-x})/(3ln(5))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 8/(5^{3x)}

\int \frac{8}{5 ^{3 x}} d x

- \frac{8 \cdot 12 5 ^{- x}}{3 ln ( 5 )} + C

Solution steps

\int \frac{8}{5 ^{3 x}} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{5 ^{3 x}} d x

Apply exponent rule:

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

\frac{1}{5 ^{3 x}} = 5^{- 3 x}

= 8 \cdot \int 5^{- 3 x} d x

Simplify

5^{- 3 x} : 12 5^{- x}

= 8 \cdot \int 12 5^{- x} d x

Apply u-substitution

: \int - 12 5^{u} d u

= 8 \cdot \int - 12 5^{u} d u

Take the constant out:

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

= 8 (- \int 12 5^{u} d u)

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

= 8 (- \frac{12 5 ^{u}}{ln ( 125 )})

Substitute back

u = - x

= 8 (- \frac{12 5 ^{- x}}{ln ( 125 )})

Simplify

8 (- \frac{12 5 ^{- x}}{ln ( 125 )}) : - \frac{8 \cdot 12 5 ^{- x}}{3 ln ( 5 )}

= - \frac{8 \cdot 12 5 ^{- x}}{3 ln ( 5 )}

Add a constant to the solution

= - \frac{8 \cdot 12 5 ^{- x}}{3 ln ( 5 )} + C

\int \frac{x ^{2} + 12 x + 12}{x ( x ^{2} - 4 )} d x

\frac{d}{d x} (\frac{8 x ^{2} + 6 x + 6}{x})

\int 16 sin^{2} (x co) s^{2} x d x

y^{^{''}} + 4 y = sec^{2} (2 x)

\int 5 tan^{3} (θ) d θ

What is the integral of 8/(5^{3x)} ?
The integral of 8/(5^{3x)} is -(8*125^{-x})/(3ln(5))+C