What is the integral from-3 to 3 of (14)/(x^2-6x-40) ?

The integral from-3 to 3 of (14)/(x^2-6x-40) is -ln(13)

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integral from-3 to 3 of (14)/(x^2-6x-40)

\int_{- 3}^{3} \frac{14}{x ^{2} - 6 x - 40} d x

- ln (13)

Decimal Notation

- 2.56494 \dots

Solution steps

\int_{- 3}^{3} \frac{14}{x ^{2} - 6 x - 40} d x

Take the constant out:

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

= 14 \cdot \int_{- 3}^{3} \frac{1}{x ^{2} - 6 x - 40} d x

Take the partial fraction of

\frac{1}{x ^{2} - 6 x - 40} : - \frac{1}{14 ( x + 4 )} + \frac{1}{14 ( x - 10 )}

= 14 \cdot \int_{- 3}^{3} - \frac{1}{14 ( x + 4 )} + \frac{1}{14 ( 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

= 14 (- \int_{- 3}^{3} \frac{1}{14 ( x + 4 )} d x + \int_{- 3}^{3} \frac{1}{14 ( x - 10 )} d x)

\int_{- 3}^{3} \frac{1}{14 ( x + 4 )} d x = \frac{1}{14} ln (7)

\int_{- 3}^{3} \frac{1}{14 ( x - 10 )} d x = \frac{1}{14} (ln (7) - ln (13))

= 14 (- \frac{1}{14} ln (7) + \frac{1}{14} (ln (7) - ln (13)))

Simplify

14 (- \frac{1}{14} ln (7) + \frac{1}{14} (ln (7) - ln (13))) : - ln (13)

= - ln (13)

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What is the integral from-3 to 3 of (14)/(x^2-6x-40) ?
The integral from-3 to 3 of (14)/(x^2-6x-40) is -ln(13)