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

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

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

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

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

Decimal Notation

- 0.46209 \dots

Solution steps

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

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

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

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

\int_{1}^{2} \frac{e ^{- \frac{1}{x}}}{x ^{2}} d x

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

\int_{2}^{5} \frac{6 - x}{x ^{3}} d x

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