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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

Add a constant to the solution

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

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