What is the integral of (4x)/((x+1)(x-1)^2) ?

The integral of (4x)/((x+1)(x-1)^2) is -ln|x+1|+ln|x-1|-2/(x-1)+C

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integral of (4x)/((x+1)(x-1)^2)

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

= 4 \cdot \int - \frac{1}{4 ( x + 1 )} + \frac{1}{4 ( x - 1 )} + \frac{1}{2 ( 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

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

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

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

= 4 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )})

Simplify

4 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )}) : - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ - \frac{2}{x - 1}

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

Add a constant to the solution

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

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What is the integral of (4x)/((x+1)(x-1)^2) ?
The integral of (4x)/((x+1)(x-1)^2) is -ln|x+1|+ln|x-1|-2/(x-1)+C