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

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

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

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

\frac{2}{3} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x + 2 ∣ - 1 + C

Solution steps

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

Multiply by the conjugate of

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

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

Apply Integration By Parts

= (x + 1) (\frac{1}{3} ln ∣ x - 1 ∣ - \frac{1}{3} ln ∣ x + 2 ∣) - \int \frac{1}{3} (ln ∣ x - 1 ∣ - ln ∣ x + 2 ∣) d x

\int \frac{1}{3} (ln ∣ x - 1 ∣ - ln ∣ x + 2 ∣) d x = \frac{1}{3} (x ln ∣ x - 1 ∣ - x ln ∣ x + 2 ∣ - ln ∣ x - 1 ∣ - 2 ln ∣ x + 2 ∣ + 3)

= (x + 1) (\frac{1}{3} ln ∣ x - 1 ∣ - \frac{1}{3} ln ∣ x + 2 ∣) - \frac{1}{3} (x ln ∣ x - 1 ∣ - x ln ∣ x + 2 ∣ - ln ∣ x - 1 ∣ - 2 ln ∣ x + 2 ∣ + 3)

Simplify

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

= \frac{2}{3} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x + 2 ∣ - 1

Add a constant to the solution

= \frac{2}{3} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x + 2 ∣ - 1 + C

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