What is the integral from 7 to 9 of y/(y^2-3y-4) ?

The integral from 7 to 9 of y/(y^2-3y-4) is (ln(5/4)+4(ln(5)-ln(3)))/5

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integral from 7 to 9 of y/(y^2-3y-4)

\int_{7}^{9} \frac{y}{y ^{2} - 3 y - 4} d y

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

Decimal Notation

0.45328 \dots

Solution steps

\int_{7}^{9} \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}^{9} \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}^{9} \frac{1}{5 ( y + 1 )} d y + \int_{7}^{9} \frac{4}{5 ( y - 4 )} d y

\int_{7}^{9} \frac{1}{5 ( y + 1 )} d y = \frac{1}{5} ln (\frac{5}{4})

\int_{7}^{9} \frac{4}{5 ( y - 4 )} d y = \frac{4}{5} (ln (5) - ln (3))

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

Simplify

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

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

\frac{d}{d t} (e^{4 t} sin (t))

f (x) = 3 x^{2} - x^{3}

\int (9 + u + u^{2}) d u

\int \frac{4}{x ^{2} ( x - 2 )} d x

\int (x^{4} + x + 9) d x

What is the integral from 7 to 9 of y/(y^2-3y-4) ?
The integral from 7 to 9 of y/(y^2-3y-4) is (ln(5/4)+4(ln(5)-ln(3)))/5