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

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

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

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

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

Solution steps

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \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

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