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

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

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

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

2 ln ∣ x - 1 ∣ - 3 ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

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

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

= 2 ln ∣ x - 1 ∣ - 3 ln ∣ x - 2 ∣

Add a constant to the solution

= 2 ln ∣ x - 1 ∣ - 3 ln ∣ x - 2 ∣ + C

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