What is the integral from 6 to 8 of y/(y^2-2y-3) ?

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

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integral from 6 to 8 of y/(y^2-2y-3)

\int_{6}^{8} \frac{y}{y ^{2} - 2 y - 3} d y

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

Decimal

0.44594 \dots

Solution steps

\int_{6}^{8} \frac{y}{y ^{2} - 2 y - 3} d y

Take the partial fraction of

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

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

\int_{6}^{8} \frac{1}{4 ( y + 1 )} d y = \frac{1}{4} (2 ln (3) - ln (7))

\int_{6}^{8} \frac{3}{4 ( y - 3 )} d y = \frac{3}{4} (ln (5) - ln (3))

= \frac{1}{4} (2 ln (3) - ln (7)) + \frac{3}{4} (ln (5) - ln (3))

Simplify

\frac{1}{4} (2 ln (3) - ln (7)) + \frac{3}{4} (ln (5) - ln (3)) : \frac{- ln ( 3 ) - ln ( 7 ) + 3 ln ( 5 )}{4}

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

\int (x^{3} - 1)^{4} x^{2} d x

n = 1 \sum \infty \frac{1}{n ( n + 2 )}

t an g e n t 1 - 4 x

n = 1 \sum \infty \frac{( 1 )}{n ^{2}}

\frac{d ^{2}}{d x ^{2}} (ln (1 + e^{x}))

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