What is the integral from-3 to 3 of (14)/(x^2-4x-45) ?

The integral from-3 to 3 of (14)/(x^2-4x-45) is -3ln(2)

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integral from-3 to 3 of (14)/(x^2-4x-45)

\int_{- 3}^{3} \frac{14}{x ^{2} - 4 x - 45} d x

- 3 ln (2)

Decimal

- 2.07944 \dots

Solution steps

\int_{- 3}^{3} \frac{14}{x ^{2} - 4 x - 45} 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} - 4 x - 45} d x

Take the partial fraction of

\frac{1}{x ^{2} - 4 x - 45} : - \frac{1}{14 ( x + 5 )} + \frac{1}{14 ( x - 9 )}

= 14 \cdot \int_{- 3}^{3} - \frac{1}{14 ( x + 5 )} + \frac{1}{14 ( 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

= 14 (- \int_{- 3}^{3} \frac{1}{14 ( x + 5 )} d x + \int_{- 3}^{3} \frac{1}{14 ( x - 9 )} d x)

\int_{- 3}^{3} \frac{1}{14 ( x + 5 )} d x = \frac{1}{7} ln (2)

\int_{- 3}^{3} \frac{1}{14 ( x - 9 )} d x = - \frac{1}{14} ln (2)

= 14 (- \frac{1}{7} ln (2) - \frac{1}{14} ln (2))

Simplify

14 (- \frac{1}{7} ln (2) - \frac{1}{14} ln (2)) : - 3 ln (2)

= - 3 ln (2)

y = (3 x + 5) x^{2}

\int \frac{x ^{2} + 5 x + 6}{x + 3} d x

y = 9 - x^{2}, y = x + 3

\int sin (6 x) d x

(- 72 sin (t))^{^{'}}

What is the integral from-3 to 3 of (14)/(x^2-4x-45) ?
The integral from-3 to 3 of (14)/(x^2-4x-45) is -3ln(2)