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

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

integral

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

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

Solution steps

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

Take the partial fraction of

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

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

\int 1 d x = x

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

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

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

Add a constant to the solution

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

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

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

in t e g r a l 5 + x

in t e g r a l (2 x + 1)^{3}

in t e g r a l 2 + \frac{1}{x ^{2}} - \frac{3}{x ^{4}}

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