What is the integral of (e^{-9x}) ?

The integral of (e^{-9x}) is -1/9 e^{-9x}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (e^{-9x})

\int (e^{- 9 x}) d x

- \frac{1}{9} e^{- 9 x} + C

Solution steps

\int e^{- 9 x} d x

Apply u-substitution

: \int - \frac{1}{9} e^{u} d u

= \int - \frac{1}{9} e^{u} d u

Take the constant out:

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

= - \frac{1}{9} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= - \frac{1}{9} e^{u}

Substitute back

u = - 9 x

= - \frac{1}{9} e^{- 9 x}

Add a constant to the solution

= - \frac{1}{9} e^{- 9 x} + C

\int 6 cos^{4} (3 x) d x

\int x^{b} d x

\int 6 (t + 2)^{2} d t

\int \frac{5 x ^{2} + 3 x + 13}{( x + 1 ) ( x ^{2} + 4 )} d x

\int 5 (2 x + 1) d x