What is the integral from-2 to 2 of (16)/(x^2-4x-60) ?

The integral from-2 to 2 of (16)/(x^2-4x-60) is -ln(2)+ln(2/3)

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integral from-2 to 2 of (16)/(x^2-4x-60)

\int_{- 2}^{2} \frac{16}{x ^{2} - 4 x - 60} d x

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

Decimal

- 1.09861 \dots

Solution steps

\int_{- 2}^{2} \frac{16}{x ^{2} - 4 x - 60} d x

Take the constant out:

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

= 16 \cdot \int_{- 2}^{2} \frac{1}{x ^{2} - 4 x - 60} d x

Take the partial fraction of

\frac{1}{x ^{2} - 4 x - 60} : - \frac{1}{16 ( x + 6 )} + \frac{1}{16 ( x - 10 )}

= 16 \cdot \int_{- 2}^{2} - \frac{1}{16 ( x + 6 )} + \frac{1}{16 ( x - 10 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 16 (- \int_{- 2}^{2} \frac{1}{16 ( x + 6 )} d x + \int_{- 2}^{2} \frac{1}{16 ( x - 10 )} d x)

\int_{- 2}^{2} \frac{1}{16 ( x + 6 )} d x = \frac{1}{16} ln (2)

\int_{- 2}^{2} \frac{1}{16 ( x - 10 )} d x = \frac{1}{16} ln (\frac{2}{3})

= 16 (- \frac{1}{16} ln (2) + \frac{1}{16} ln (\frac{2}{3}))

Simplify

16 (- \frac{1}{16} ln (2) + \frac{1}{16} ln (\frac{2}{3})) : - ln (2) + ln (\frac{2}{3})

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

What is the integral from-2 to 2 of (16)/(x^2-4x-60) ?
The integral from-2 to 2 of (16)/(x^2-4x-60) is -ln(2)+ln(2/3)