What is the integral from-3 to 3 of (16)/(x^2-2x-63) ?

The integral from-3 to 3 of (16)/(x^2-2x-63) is -ln(5/2)-ln(2)

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integral from-3 to 3 of (16)/(x^2-2x-63)

\int_{- 3}^{3} \frac{16}{x ^{2} - 2 x - 63} d x

- ln (\frac{5}{2}) - ln (2)

Decimal

- 1.60943 \dots

Solution steps

\int_{- 3}^{3} \frac{16}{x ^{2} - 2 x - 63} d x

Take the constant out:

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

= 16 \cdot \int_{- 3}^{3} \frac{1}{x ^{2} - 2 x - 63} d x

Take the partial fraction of

\frac{1}{x ^{2} - 2 x - 63} : - \frac{1}{16 ( x + 7 )} + \frac{1}{16 ( x - 9 )}

= 16 \cdot \int_{- 3}^{3} - \frac{1}{16 ( x + 7 )} + \frac{1}{16 ( x - 9 )} 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_{- 3}^{3} \frac{1}{16 ( x + 7 )} d x + \int_{- 3}^{3} \frac{1}{16 ( x - 9 )} d x)

\int_{- 3}^{3} \frac{1}{16 ( x + 7 )} d x = \frac{1}{16} ln (\frac{5}{2})

\int_{- 3}^{3} \frac{1}{16 ( x - 9 )} d x = - \frac{1}{16} ln (2)

= 16 (- \frac{1}{16} ln (\frac{5}{2}) - \frac{1}{16} ln (2))

Simplify

16 (- \frac{1}{16} ln (\frac{5}{2}) - \frac{1}{16} ln (2)) : - ln (\frac{5}{2}) - ln (2)

= - ln (\frac{5}{2}) - ln (2)

What is the integral from-3 to 3 of (16)/(x^2-2x-63) ?
The integral from-3 to 3 of (16)/(x^2-2x-63) is -ln(5/2)-ln(2)