What is the integral of 2/(x^2(x+1)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 2/(x^2(x+1))

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

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

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

= 2 \cdot \int - \frac{1}{x} + \frac{1}{x ^{2}} + \frac{1}{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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

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

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

Add a constant to the solution

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

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