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

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

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

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

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