What is the integral from-2 to 2 of (10)/(x^2-4x-21) ?

The integral from-2 to 2 of (10)/(x^2-4x-21) is -2ln(3)

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integral from-2 to 2 of (10)/(x^2-4x-21)

\int_{- 2}^{2} \frac{10}{x ^{2} - 4 x - 21} d x

- 2 ln (3)

Solution steps

\int_{- 2}^{2} \frac{10}{x ^{2} - 4 x - 21} d x

Take the constant out:

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

= 10 \cdot \int_{- 2}^{2} \frac{1}{x ^{2} - 4 x - 21} d x

Take the partial fraction of

\frac{1}{x ^{2} - 4 x - 21} : - \frac{1}{10 ( x + 3 )} + \frac{1}{10 ( x - 7 )}

= 10 \cdot \int_{- 2}^{2} - \frac{1}{10 ( x + 3 )} + \frac{1}{10 ( x - 7 )} 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

= 10 (- \int_{- 2}^{2} \frac{1}{10 ( x + 3 )} d x + \int_{- 2}^{2} \frac{1}{10 ( x - 7 )} d x)

\int_{- 2}^{2} \frac{1}{10 ( x + 3 )} d x = \frac{1}{10} ln (5)

\int_{- 2}^{2} \frac{1}{10 ( x - 7 )} d x = \frac{1}{10} (ln (5) - 2 ln (3))

= 10 (- \frac{1}{10} ln (5) + \frac{1}{10} (ln (5) - 2 ln (3)))

Simplify

10 (- \frac{1}{10} ln (5) + \frac{1}{10} (ln (5) - 2 ln (3))) : - 2 ln (3)

= - 2 ln (3)

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What is the integral from-2 to 2 of (10)/(x^2-4x-21) ?
The integral from-2 to 2 of (10)/(x^2-4x-21) is -2ln(3)