What is the integral of 6/(x(x+2)) ?

The integral of 6/(x(x+2)) is 6(1/2 ln|x|-1/2 ln|x+2|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 6/(x(x+2))

\int \frac{6}{x ( x + 2 )} d x

6 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 2 ∣) + C

Solution steps

\int \frac{6}{x ( x + 2 )} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{x ( x + 2 )} d x

Take the partial fraction of

\frac{1}{x ( x + 2 )} : \frac{1}{2 x} - \frac{1}{2 ( x + 2 )}

= 6 \cdot \int \frac{1}{2 x} - \frac{1}{2 ( x + 2 )} 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

= 6 (\int \frac{1}{2 x} d x - \int \frac{1}{2 ( x + 2 )} d x)

\int \frac{1}{2 x} d x = \frac{1}{2} ln ∣ x ∣

\int \frac{1}{2 ( x + 2 )} d x = \frac{1}{2} ln ∣ x + 2 ∣

= 6 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 2 ∣)

Add a constant to the solution

= 6 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 2 ∣) + C

x \to 3 + lim (\frac{∣ x - 2 ∣}{x - 2})

\frac{\partial}{\partial x} (56 x^{2} - 8 y^{2} - 16 x - 31 + 1 - 8 x)

f (x) = 1 + x

\int \frac{1}{x ^{1 + c}} d x

\int \frac{- 5 x ^{2}}{x ^{4} - 1} d x

What is the integral of 6/(x(x+2)) ?
The integral of 6/(x(x+2)) is 6(1/2 ln|x|-1/2 ln|x+2|)+C