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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

integral

(4 + x)^{- 1}

\frac{x}{4 + x} + ln ∣ 4 + x ∣ + \frac{4}{4 + x} + C

Solution steps

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

Apply Integration By Parts

= \frac{x}{4 + x} - \int - \frac{x}{( 4 + x ) ^{2}} d x

\int - \frac{x}{( 4 + x ) ^{2}} d x = - ln ∣ 4 + x ∣ - \frac{4}{4 + x}

= \frac{x}{4 + x} - (- ln ∣ 4 + x ∣ - \frac{4}{4 + x})

Simplify

= \frac{x}{4 + x} + ln ∣ 4 + x ∣ + \frac{4}{4 + x}

Add a constant to the solution

= \frac{x}{4 + x} + ln ∣ 4 + x ∣ + \frac{4}{4 + x} + C

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