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

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

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

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

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

Solution steps

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

Expand

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

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

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

Simplify

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

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

Add a constant to the solution

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

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