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

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

Popular >

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

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

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

Solution steps

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

Take the partial fraction of

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

\int_{1}^{6} x^{3} ln (4 x) d x

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

\int \frac{( ln ( x ) ) ^{22}}{x} d x

\int x (15 x + 6) d x

\int \frac{1}{( x - 2 ) ( x ^{2} + 4 )} d x

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