What is the integral from-1 to 5 of (8y)/(y^2-4y-12) ?

The integral from-1 to 5 of (8y)/(y^2-4y-12) is -4ln(7)

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integral from-1 to 5 of (8y)/(y^2-4y-12)

\int_{- 1}^{5} \frac{8 y}{y ^{2} - 4 y - 12} d y

- 4 ln (7)

Decimal Notation

- 7.78364 \dots

Solution steps

\int_{- 1}^{5} \frac{8 y}{y ^{2} - 4 y - 12} d y

Take the constant out:

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

= 8 \cdot \int_{- 1}^{5} \frac{y}{y ^{2} - 4 y - 12} d y

Take the partial fraction of

\frac{y}{y ^{2} - 4 y - 12} : \frac{1}{4 ( y + 2 )} + \frac{3}{4 ( y - 6 )}

= 8 \cdot \int_{- 1}^{5} \frac{1}{4 ( y + 2 )} + \frac{3}{4 ( y - 6 )} 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

= 8 (\int_{- 1}^{5} \frac{1}{4 ( y + 2 )} d y + \int_{- 1}^{5} \frac{3}{4 ( y - 6 )} d y)

\int_{- 1}^{5} \frac{1}{4 ( y + 2 )} d y = \frac{1}{4} ln (7)

\int_{- 1}^{5} \frac{3}{4 ( y - 6 )} d y = - \frac{3}{4} ln (7)

= 8 (\frac{1}{4} ln (7) - \frac{3}{4} ln (7))

Simplify

8 (\frac{1}{4} ln (7) - \frac{3}{4} ln (7)) : - 4 ln (7)

= - 4 ln (7)

\int 16 csc (8 x) sec (8 x) d x

x^{2} + 2 x + 1, - 1, 7

y = 8 x, y = x^{2} - 9

\frac{d}{d x} (\frac{400}{1 + 39 e ^{- 0.16 x}})

\int \frac{3 + t + t ^{2}}{t} d t

What is the integral from-1 to 5 of (8y)/(y^2-4y-12) ?
The integral from-1 to 5 of (8y)/(y^2-4y-12) is -4ln(7)