What is the integral from 8 to 13 of y/(y^2-4y-5) ?

The integral from 8 to 13 of y/(y^2-4y-5) is (-7ln(3)+15ln(2)+ln(14))/6

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integral from 8 to 13 of y/(y^2-4y-5)

\int_{8}^{13} \frac{y}{y ^{2} - 4 y - 5} d y

\frac{- 7 ln ( 3 ) + 15 ln ( 2 ) + ln ( 14 )}{6}

Solution steps

\int_{8}^{13} \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}^{13} \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}^{13} \frac{1}{6 ( y + 1 )} d y + \int_{8}^{13} \frac{5}{6 ( y - 5 )} d y

\int_{8}^{13} \frac{1}{6 ( y + 1 )} d y = \frac{1}{6} (ln (14) - 2 ln (3))

\int_{8}^{13} \frac{5}{6 ( y - 5 )} d y = \frac{5}{6} (3 ln (2) - ln (3))

= \frac{1}{6} (ln (14) - 2 ln (3)) + \frac{5}{6} (3 ln (2) - ln (3))

Simplify

\frac{1}{6} (ln (14) - 2 ln (3)) + \frac{5}{6} (3 ln (2) - ln (3)) : \frac{- 7 ln ( 3 ) + 15 ln ( 2 ) + ln ( 14 )}{6}

= \frac{- 7 ln ( 3 ) + 15 ln ( 2 ) + ln ( 14 )}{6}

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 e^{2 t}

\frac{d}{d x} (- 3 x + 7)

\frac{\partial}{\partial x} (3 e^{x y + 4})

\frac{d y}{d x} = x y + 5 x + 2 y + 10

13 sin (2 t)

What is the integral from 8 to 13 of y/(y^2-4y-5) ?
The integral from 8 to 13 of y/(y^2-4y-5) is (-7ln(3)+15ln(2)+ln(14))/6