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

The integral of ((4-2x))/((x+1)(x-1)^2) is 3/2 ln|x+1|-3/2 ln|x-1|-1/(x-1)+C

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int_{1}^{\infty} \frac{3}{x ^{2}} d x

x \to - 4 lim (x^{2} - 8)

\int \frac{1}{( 3 x + 5 ) ^{3}} d x

f (x) = 2 x^{3} - 9 x^{2} + 12 x - 6

\frac{d y}{d t} = k (y - 20)

What is the integral of ((4-2x))/((x+1)(x-1)^2) ?
The integral of ((4-2x))/((x+1)(x-1)^2) is 3/2 ln|x+1|-3/2 ln|x-1|-1/(x-1)+C