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

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

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

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

- ln ∣ x - 1 ∣ + ln ∣ x ∣ + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

= - ln ∣ x - 1 ∣ + ln ∣ x ∣

Add a constant to the solution

= - ln ∣ x - 1 ∣ + ln ∣ x ∣ + C

x \to 0 lim (x^{s i n (3 x)})

Y = (3 - X^{2}) (X^{3} - X + 1), (2, 1)

x \to + π lim (\frac{x}{4} - \frac{π}{2})

\frac{d y}{d x} = 2 x y^{2} + 32 x

\frac{d}{d x} (\frac{3}{2} x^{\frac{2}{3}})

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