What is the integral from 0 to 1 of 1/(x^2-2x-8) ?

The integral from 0 to 1 of 1/(x^2-2x-8) is -1/6 ln(2)

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integral from 0 to 1 of 1/(x^2-2x-8)

\int_{0}^{1} \frac{1}{x ^{2} - 2 x - 8} d x

- \frac{1}{6} ln (2)

Decimal

- 0.11552 \dots

Solution steps

\int_{0}^{1} \frac{1}{x ^{2} - 2 x - 8} d x

Take the partial fraction of

\frac{1}{x ^{2} - 2 x - 8} : - \frac{1}{6 ( x + 2 )} + \frac{1}{6 ( x - 4 )}

= \int_{0}^{1} - \frac{1}{6 ( x + 2 )} + \frac{1}{6 ( x - 4 )} 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

= - \int_{0}^{1} \frac{1}{6 ( x + 2 )} d x + \int_{0}^{1} \frac{1}{6 ( x - 4 )} d x

\int_{0}^{1} \frac{1}{6 ( x + 2 )} d x = \frac{1}{6} (ln (3) - ln (2))

\int_{0}^{1} \frac{1}{6 ( x - 4 )} d x = \frac{1}{6} (ln (3) - 2 ln (2))

= - \frac{1}{6} (ln (3) - ln (2)) + \frac{1}{6} (ln (3) - 2 ln (2))

Simplify

- \frac{1}{6} (ln (3) - ln (2)) + \frac{1}{6} (ln (3) - 2 ln (2)) : - \frac{1}{6} ln (2)

= - \frac{1}{6} ln (2)

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What is the integral from 0 to 1 of 1/(x^2-2x-8) ?
The integral from 0 to 1 of 1/(x^2-2x-8) is -1/6 ln(2)