What is the integral from 2 to 4 of x/((x+1)(x+4)) ?

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

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integral from 2 to 4 of x/((x+1)(x+4))

\int_{2}^{4} \frac{x}{( x + 1 ) ( x + 4 )} d x

\frac{- ln ( 5 ) + ln ( 3 ) + 4 ln ( \frac{4}{3} )}{3}

Decimal Notation

0.21330 \dots

Solution steps

\int_{2}^{4} \frac{x}{( x + 1 ) ( x + 4 )} d x

Take the partial fraction of

\frac{x}{( x + 1 ) ( x + 4 )} : - \frac{1}{3 ( x + 1 )} + \frac{4}{3 ( x + 4 )}

= \int_{2}^{4} - \frac{1}{3 ( x + 1 )} + \frac{4}{3 ( x + 4 )} 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}^{4} \frac{1}{3 ( x + 1 )} d x + \int_{2}^{4} \frac{4}{3 ( x + 4 )} d x

\int_{2}^{4} \frac{1}{3 ( x + 1 )} d x = \frac{1}{3} (ln (5) - ln (3))

\int_{2}^{4} \frac{4}{3 ( x + 4 )} d x = \frac{4}{3} ln (\frac{4}{3})

= - \frac{1}{3} (ln (5) - ln (3)) + \frac{4}{3} ln (\frac{4}{3})

Simplify

- \frac{1}{3} (ln (5) - ln (3)) + \frac{4}{3} ln (\frac{4}{3}) : \frac{- ln ( 5 ) + ln ( 3 ) + 4 ln ( \frac{4}{3} )}{3}

= \frac{- ln ( 5 ) + ln ( 3 ) + 4 ln ( \frac{4}{3} )}{3}

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