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

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

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

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

ln ∣ x + 1 ∣ + \frac{2}{x + 1} + C

Solution steps

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

Expand

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

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

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

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

= ln ∣ x + 1 ∣ + \frac{1}{x + 1} - (- \frac{1}{x + 1})

Simplify

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

= ln ∣ x + 1 ∣ + \frac{2}{x + 1}

Add a constant to the solution

= ln ∣ x + 1 ∣ + \frac{2}{x + 1} + C

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