What is the integral from 5 to 10 of y/(y^2-y-2) ?

The integral from 5 to 10 of y/(y^2-y-2) is 2ln(2)+(ln(11)-ln(6)-2ln(3))/3

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integral from 5 to 10 of y/(y^2-y-2)

\int_{5}^{10} \frac{y}{y ^{2} - y - 2} d y

2 ln (2) + \frac{ln ( 11 ) - ln ( 6 ) - 2 ln ( 3 )}{3}

Decimal

0.85593 \dots

Solution steps

\int_{5}^{10} \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}^{10} \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}^{10} \frac{1}{3 ( y + 1 )} d y + \int_{5}^{10} \frac{2}{3 ( y - 2 )} d y

\int_{5}^{10} \frac{1}{3 ( y + 1 )} d y = \frac{1}{3} (ln (11) - ln (6))

\int_{5}^{10} \frac{2}{3 ( y - 2 )} d y = 2 ln (2) - \frac{2}{3} ln (3)

= \frac{1}{3} (ln (11) - ln (6)) + 2 ln (2) - \frac{2}{3} ln (3)

Simplify

\frac{1}{3} (ln (11) - ln (6)) + 2 ln (2) - \frac{2}{3} ln (3) : 2 ln (2) + \frac{ln ( 11 ) - ln ( 6 ) - 2 ln ( 3 )}{3}

= 2 ln (2) + \frac{ln ( 11 ) - ln ( 6 ) - 2 ln ( 3 )}{3}

8 sec (x)

x \to - a lim (x)

\frac{d}{d x} ((\frac{x}{x + 5})^{4})

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

What is the integral from 5 to 10 of y/(y^2-y-2) ?
The integral from 5 to 10 of y/(y^2-y-2) is 2ln(2)+(ln(11)-ln(6)-2ln(3))/3