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

The integral of 1/(e^{-3x)} is 1/3 e^{3x}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(e^{-3x)}

\int \frac{1}{e ^{- 3 x}} d x

\frac{1}{3} e^{3 x} + C

Solution steps

\int \frac{1}{e ^{- 3 x}} d x

Apply exponent rule:

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

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

= \int e^{3 x} d x

Apply u-substitution

= \int e^{u} \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{3} e^{u}

Substitute back

u = 3 x

= \frac{1}{3} e^{3 x}

Add a constant to the solution

= \frac{1}{3} e^{3 x} + C

What is the integral of 1/(e^{-3x)} ?
The integral of 1/(e^{-3x)} is 1/3 e^{3x}+C