What is the integral of ((x+6))/(x+1) ?

The integral of ((x+6))/(x+1) is 5ln|x+1|+x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((x+6))/(x+1)

\int \frac{( x + 6 )}{x + 1} d x

5 ln ∣ x + 1 ∣ + x + C

Solution steps

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

\frac{x + 6}{x + 1} = \frac{5}{x + 1} + 1

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

\int \frac{5}{x + 1} d x = 5 ln ∣ x + 1 ∣

\int 1 d x = x

= 5 ln ∣ x + 1 ∣ + x

Add a constant to the solution

= 5 ln ∣ x + 1 ∣ + x + C

\frac{\partial}{\partial y} (x^{4} - x^{2} y)

\int_{- 1}^{2} (x^{3} - x + 2) d x

\frac{d}{d x} ((10 - 5 x)^{4})

f (x) = (cot (x))^{4}

g (u) = (\frac{u ^{3} - 1}{u ^{3} + 1})^{8}

What is the integral of ((x+6))/(x+1) ?
The integral of ((x+6))/(x+1) is 5ln|x+1|+x+C