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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

\int 2 x y + 3 x^{2} d x

\int_{0}^{2} (2 x - 1) d x

\int arctan (a x) d x

x \to 3 lim (5^{2})

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