What is the integral of (e^{4x}-x^{-1}) ?

The integral of (e^{4x}-x^{-1}) is 1/4 e^{4x}-ln|x|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (e^{4x}-x^{-1})

\int (e^{4 x} - x^{- 1}) d x

\frac{1}{4} e^{4 x} - ln ∣ x ∣ + C

Solution steps

\int e^{4 x} - x^{- 1} 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 e^{4 x} d x - \int x^{- 1} d x

\int e^{4 x} d x = \frac{1}{4} e^{4 x}

\int x^{- 1} d x = ln ∣ x ∣

= \frac{1}{4} e^{4 x} - ln ∣ x ∣

Add a constant to the solution

= \frac{1}{4} e^{4 x} - ln ∣ x ∣ + C

\int x arctan (x) ln (1 + x^{2}) d x

\int \frac{1}{x ^{2} + 8 x + 7} d x

\int s^{2} d s

\int \frac{2 x}{5 x + 1} d x

\int x^{2} + 9 x d x

What is the integral of (e^{4x}-x^{-1}) ?
The integral of (e^{4x}-x^{-1}) is 1/4 e^{4x}-ln|x|+C