What is the integral from-1 to 3 of (6y)/(y^2-2y-8) ?

The integral from-1 to 3 of (6y)/(y^2-2y-8) is -2ln(5)

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integral from-1 to 3 of (6y)/(y^2-2y-8)

\int_{- 1}^{3} \frac{6 y}{y ^{2} - 2 y - 8} d y

- 2 ln (5)

Decimal Notation

- 3.21887 \dots

Solution steps

\int_{- 1}^{3} \frac{6 y}{y ^{2} - 2 y - 8} d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int_{- 1}^{3} \frac{y}{y ^{2} - 2 y - 8} d y

Take the partial fraction of

\frac{y}{y ^{2} - 2 y - 8} : \frac{1}{3 ( y + 2 )} + \frac{2}{3 ( y - 4 )}

= 6 \cdot \int_{- 1}^{3} \frac{1}{3 ( y + 2 )} + \frac{2}{3 ( 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

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

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

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

= 6 (\frac{1}{3} ln (5) - \frac{2}{3} ln (5))

Simplify

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

= - 2 ln (5)

\int \frac{u - 1}{u} d u

y = \frac{1}{x}, y = \frac{1}{3}, 1, 3

\int \frac{1}{x ^{2} + 2 x - 3} d x

\frac{d}{d x} (\frac{x ^{6}}{6})

y = cot^{2} (sin (θ))

What is the integral from-1 to 3 of (6y)/(y^2-2y-8) ?
The integral from-1 to 3 of (6y)/(y^2-2y-8) is -2ln(5)