What is the integral of (3^{-x}) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3^{-x})

\int (3^{- x}) d x

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

Solution steps

\int 3^{- x} d x

Apply u-substitution

: \int - 3^{u} d u

= \int - 3^{u} d u

Take the constant out:

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

= - \int 3^{u} d u

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

= - \frac{3 ^{u}}{ln ( 3 )}

Substitute back

u = - x

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

Add a constant to the solution

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

\frac{d y}{d x}, y = x^{2} + 3

\frac{d}{d x} (x sin (\frac{π}{x}))

f (x) = tan (ln (x))

\int_{- 3}^{5} (2 y - (y^{2} - 15)) d y

2 x^{\frac{3}{2}}

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