What is the integral from 0 to 1 of 1/((x+1)(x-2)) ?

The integral from 0 to 1 of 1/((x+1)(x-2)) is -2/3 ln(2)

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integral from 0 to 1 of 1/((x+1)(x-2))

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

- \frac{2}{3} ln (2)

Decimal Notation

- 0.46209 \dots

Solution steps

\int_{0}^{1} \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_{0}^{1} - \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_{0}^{1} \frac{1}{3 ( x + 1 )} d x + \int_{0}^{1} \frac{1}{3 ( x - 2 )} d x

\int_{0}^{1} \frac{1}{3 ( x + 1 )} d x = \frac{1}{3} ln (2)

\int_{0}^{1} \frac{1}{3 ( x - 2 )} d x = - \frac{1}{3} ln (2)

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

Simplify

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

= - \frac{2}{3} ln (2)

\int \frac{x ^{2} - a ^{2}}{x} d x

\frac{\partial}{\partial y} (cos (x^{2}) sin (yz))

\frac{d y}{d x} = \frac{y ^{2} + 4 x x ^{2} + y ^{2}}{x y}

f (x) = \frac{1}{x ^{5}}

f (x) = (x^{3} - 3 x) (2 x^{2} + 3 x + 5)

What is the integral from 0 to 1 of 1/((x+1)(x-2)) ?
The integral from 0 to 1 of 1/((x+1)(x-2)) is -2/3 ln(2)