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

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