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

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

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

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

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

Decimal Notation

0.59725 \dots

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Simplify

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

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

e^{x}, e^{5 x}, 0, 1

\frac{d}{d x} (1 - sin (2 x))

\frac{\partial}{\partial x} (z^{3} ln (y x))

2 \cdot ((t - 1)^{2}) \cdot 1 ((t - 1))

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