What is the integral from 4 to 8 of y/(y^2-2y-3) ?

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

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integral from 4 to 8 of y/(y^2-2y-3)

\int_{4}^{8} \frac{y}{y ^{2} - 2 y - 3} d y

\frac{2 ln ( 3 ) + 2 ln ( 5 )}{4}

Decimal

1.35402 \dots

Solution steps

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

\int_{4}^{8} \frac{1}{4 ( y + 1 )} d y = \frac{1}{4} (2 ln (3) - ln (5))

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

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

Simplify

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

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

x^{3} - 2 x^{2} + 3, 3 x^{2} + 5 x - 22

\int - e^{6 t} d t

\frac{4}{s ( 5 s + 1 )}

x y^{^{'}} = 2 y + 3 x

\frac{d}{d x} (\frac{1}{5 x - 6})

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