What is the integral of (-7x+1)/((x+1)(x-1)^2) ?

The integral of (-7x+1)/((x+1)(x-1)^2) is 2ln|x+1|-2ln|x-1|+3/(x-1)+C

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integral of (-7x+1)/((x+1)(x-1)^2)

\int \frac{- 7 x + 1}{( x + 1 ) ( x - 1 ) ^{2}} d x

2 ln ∣ x + 1 ∣ - 2 ln ∣ x - 1 ∣ + \frac{3}{x - 1} + C

Solution steps

\int \frac{- 7 x + 1}{( x + 1 ) ( x - 1 ) ^{2}} d x

Expand

\frac{- 7 x + 1}{( x + 1 ) ( x - 1 ) ^{2}} : - \frac{7 x}{( x + 1 ) ( x - 1 ) ^{2}} + \frac{1}{( x + 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{7 x}{( x + 1 ) ( x - 1 ) ^{2}} d x + \int \frac{1}{( x + 1 ) ( x - 1 ) ^{2}} d x

\int \frac{7 x}{( x + 1 ) ( x - 1 ) ^{2}} d x = 7 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )})

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

= - 7 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )}) + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )}

Simplify

- 7 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )}) + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} : 2 ln ∣ x + 1 ∣ - 2 ln ∣ x - 1 ∣ + \frac{3}{x - 1}

= 2 ln ∣ x + 1 ∣ - 2 ln ∣ x - 1 ∣ + \frac{3}{x - 1}

Add a constant to the solution

= 2 ln ∣ x + 1 ∣ - 2 ln ∣ x - 1 ∣ + \frac{3}{x - 1} + C

What is the integral of (-7x+1)/((x+1)(x-1)^2) ?
The integral of (-7x+1)/((x+1)(x-1)^2) is 2ln|x+1|-2ln|x-1|+3/(x-1)+C