What is the integral from 8 to 14 of y/(y^2-4y-5) ?

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

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integral from 8 to 14 of y/(y^2-4y-5)

\int_{8}^{14} \frac{y}{y ^{2} - 4 y - 5} d y

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

Decimal Notation

1.00064 \dots

Solution steps

\int_{8}^{14} \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}^{14} \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}^{14} \frac{1}{6 ( y + 1 )} d y + \int_{8}^{14} \frac{5}{6 ( y - 5 )} d y

\int_{8}^{14} \frac{1}{6 ( y + 1 )} d y = \frac{1}{6} ln (\frac{5}{3})

\int_{8}^{14} \frac{5}{6 ( y - 5 )} d y = \frac{5}{6} ln (3)

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

Simplify

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

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

y^{^{'}} = e^{- 2 x}

26 e^{x^{2}} x

f (x) = x^{2} + 1

7 y^{^{''}} + 5 y^{^{'}} + 7 y = 0

h \to 0 lim (\frac{\frac{4}{2 + h} - 2}{h})

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