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

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

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

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

- ln ∣ x - 1 ∣ + ln ∣ 3 x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

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

= \int - \frac{1}{x - 1} + \frac{3}{3 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{3}{3 x - 2} d x

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

\int \frac{3}{3 x - 2} d x = ln ∣ 3 x - 2 ∣

= - ln ∣ x - 1 ∣ + ln ∣ 3 x - 2 ∣

Add a constant to the solution

= - ln ∣ x - 1 ∣ + ln ∣ 3 x - 2 ∣ + C

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