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_{6}^{8} (\frac{x ^{2} + 6}{x ^{2}}) d x

\int \frac{x ^{3} - x ^{2} + x - 1}{x ^{2} + 9} d x

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

x \to 2 lim (lo g_{4} (x - 2))

\frac{d y}{d x} + 3 x^{2} y = x^{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