What is the integral of (3x-2)/((x+1)^2(x-1)) ?

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

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integral of (3x-2)/((x+1)^2(x-1))

\int \frac{3 x - 2}{( x + 1 ) ^{2} ( x - 1 )} d x

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

Solution steps

\int \frac{3 x - 2}{( x + 1 ) ^{2} ( x - 1 )} d x

Expand

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

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

\int \frac{3 x}{( x + 1 ) ^{2} ( x - 1 )} d x = 3 (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣)

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

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

Simplify

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

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

Add a constant to the solution

= - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ - \frac{3}{2 ( x + 1 )} - \frac{1}{x + 1} + C

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