What is the integral from 8 to 11 of y/(y^2-4y-5) ?

The integral from 8 to 11 of y/(y^2-4y-5) is (ln(4/3)+5ln(2))/6

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integral from 8 to 11 of y/(y^2-4y-5)

\int_{8}^{11} \frac{y}{y ^{2} - 4 y - 5} d y

\frac{ln ( \frac{4}{3} ) + 5 ln ( 2 )}{6}

Decimal

0.62556 \dots

Solution steps

\int_{8}^{11} \frac{y}{y ^{2} - 4 y - 5} d y

Take the partial fraction of

\frac{y}{y ^{2} - 4 y - 5} : \frac{1}{6 ( y + 1 )} + \frac{5}{6 ( y - 5 )}

= \int_{8}^{11} \frac{1}{6 ( y + 1 )} + \frac{5}{6 ( y - 5 )} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{8}^{11} \frac{1}{6 ( y + 1 )} d y + \int_{8}^{11} \frac{5}{6 ( y - 5 )} d y

\int_{8}^{11} \frac{1}{6 ( y + 1 )} d y = \frac{1}{6} ln (\frac{4}{3})

\int_{8}^{11} \frac{5}{6 ( y - 5 )} d y = \frac{5}{6} ln (2)

= \frac{1}{6} ln (\frac{4}{3}) + \frac{5}{6} ln (2)

Simplify

\frac{1}{6} ln (\frac{4}{3}) + \frac{5}{6} ln (2) : \frac{ln ( \frac{4}{3} ) + 5 ln ( 2 )}{6}

= \frac{ln ( \frac{4}{3} ) + 5 ln ( 2 )}{6}

What is the integral from 8 to 11 of y/(y^2-4y-5) ?
The integral from 8 to 11 of y/(y^2-4y-5) is (ln(4/3)+5ln(2))/6