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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

x \to 1 lim (x + 1)

\int \frac{2}{x ^{2} + 11 x + 18} d x

\int 5 cos (3 x) - 4 sin (5 x) d x

lo g_{5} (x)

\frac{1}{3} x^{\frac{1}{2}} - x^{\frac{3}{2}}, x = 0, x = \frac{1}{3}

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