What is the integral from 2 to 5 of 1/((x-1)(x+2)) ?

The integral from 2 to 5 of 1/((x-1)(x+2)) is (4ln(2)-ln(7))/3

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integral from 2 to 5 of 1/((x-1)(x+2))

\int_{2}^{5} \frac{1}{( x - 1 ) ( x + 2 )} d x

\frac{4 ln ( 2 ) - ln ( 7 )}{3}

Decimal Notation

0.27555 \dots

Solution steps

\int_{2}^{5} \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_{2}^{5} \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_{2}^{5} \frac{1}{3 ( x - 1 )} d x - \int_{2}^{5} \frac{1}{3 ( x + 2 )} d x

\int_{2}^{5} \frac{1}{3 ( x - 1 )} d x = \frac{2}{3} ln (2)

\int_{2}^{5} \frac{1}{3 ( x + 2 )} d x = \frac{1}{3} (ln (7) - 2 ln (2))

= \frac{2}{3} ln (2) - \frac{1}{3} (ln (7) - 2 ln (2))

Simplify

\frac{2}{3} ln (2) - \frac{1}{3} (ln (7) - 2 ln (2)) : \frac{4 ln ( 2 ) - ln ( 7 )}{3}

= \frac{4 ln ( 2 ) - ln ( 7 )}{3}

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What is the integral from 2 to 5 of 1/((x-1)(x+2)) ?
The integral from 2 to 5 of 1/((x-1)(x+2)) is (4ln(2)-ln(7))/3