What is the integral from 7 to 13 of y/(y^2-3y-4) ?

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

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integral from 7 to 13 of y/(y^2-3y-4)

\int_{7}^{13} \frac{y}{y ^{2} - 3 y - 4} d y

\frac{ln ( \frac{7}{4} ) + 4 ln ( 3 )}{5}

Decimal Notation

0.99081 \dots

Solution steps

\int_{7}^{13} \frac{y}{y ^{2} - 3 y - 4} d y

Take the partial fraction of

\frac{y}{y ^{2} - 3 y - 4} : \frac{1}{5 ( y + 1 )} + \frac{4}{5 ( y - 4 )}

= \int_{7}^{13} \frac{1}{5 ( y + 1 )} + \frac{4}{5 ( y - 4 )} 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_{7}^{13} \frac{1}{5 ( y + 1 )} d y + \int_{7}^{13} \frac{4}{5 ( y - 4 )} d y

\int_{7}^{13} \frac{1}{5 ( y + 1 )} d y = \frac{1}{5} ln (\frac{7}{4})

\int_{7}^{13} \frac{4}{5 ( y - 4 )} d y = \frac{4}{5} ln (3)

= \frac{1}{5} ln (\frac{7}{4}) + \frac{4}{5} ln (3)

Simplify

\frac{1}{5} ln (\frac{7}{4}) + \frac{4}{5} ln (3) : \frac{ln ( \frac{7}{4} ) + 4 ln ( 3 )}{5}

= \frac{ln ( \frac{7}{4} ) + 4 ln ( 3 )}{5}

f^{^{'''}} (x) = 6 (x^{- 2})

\frac{s + 10}{s ^{2} - s - 2}

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

\int e^{x} - e^{2 x} d x

\frac{1}{s - 2}

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