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

The integral from 2 to 5 of 1/((x+2)(x+5)) is (2ln(7)-2ln(2)-ln(10))/3

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

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

\frac{2 ln ( 7 ) - 2 ln ( 2 ) - ln ( 10 )}{3}

Decimal

0.06764 \dots

Solution steps

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

Take the partial fraction of

\frac{1}{( x + 2 ) ( x + 5 )} : \frac{1}{3 ( x + 2 )} - \frac{1}{3 ( x + 5 )}

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

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

\int_{2}^{5} \frac{1}{3 ( x + 5 )} d x = \frac{1}{3} (ln (10) - ln (7))

= \frac{1}{3} (ln (7) - 2 ln (2)) - \frac{1}{3} (ln (10) - ln (7))

Simplify

\frac{1}{3} (ln (7) - 2 ln (2)) - \frac{1}{3} (ln (10) - ln (7)) : \frac{2 ln ( 7 ) - 2 ln ( 2 ) - ln ( 10 )}{3}

= \frac{2 ln ( 7 ) - 2 ln ( 2 ) - ln ( 10 )}{3}

What is the integral from 2 to 5 of 1/((x+2)(x+5)) ?
The integral from 2 to 5 of 1/((x+2)(x+5)) is (2ln(7)-2ln(2)-ln(10))/3