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

The integral of 1/x-e^{-x} is ln|x|+e^{-x}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/x-e^{-x}

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

ln ∣ x ∣ + e^{- x} + C

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int e^{- x} d x = - e^{- x}

= ln ∣ x ∣ - (- e^{- x})

Simplify

= ln ∣ x ∣ + e^{- x}

Add a constant to the solution

= ln ∣ x ∣ + e^{- x} + C

\int \frac{1}{4 x + 6 x} d x

n = 1 \sum \infty \frac{1}{2 n - 1}

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

f (x) = \frac{1}{x}, at x = 3

x \to + 0 lim ((1 + 6 x)^{\frac{4}{x}})

What is the integral of 1/x-e^{-x} ?
The integral of 1/x-e^{-x} is ln|x|+e^{-x}+C