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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

= \int \frac{2}{x - 1} - \frac{2}{x} - \frac{1}{x ^{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

= \int \frac{2}{x - 1} d x - \int \frac{2}{x} d x - \int \frac{1}{x ^{2}} d x

\int \frac{2}{x - 1} d x = 2 ln ∣ x - 1 ∣

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

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

Simplify

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

Add a constant to the solution

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

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