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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

integral

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

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

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

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

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

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

Add a constant to the solution

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

in t e g r a l \frac{1}{( x ( x + 2 ) )}

in t e g r a l 5^{- x}

in t e g r a l \frac{y}{( x ^{2} + y ^{2} )}

in t e g r a l \frac{x ^{2}}{( ( x - 2 ) ( x ^{2} + 1 ) )}

in t e g r a l 1 + y^{2}

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