What is the integral from 0 to 1 of 1/(x^2-2x-3) ?

The integral from 0 to 1 of 1/(x^2-2x-3) is -1/4 ln(3)

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integral from 0 to 1 of 1/(x^2-2x-3)

\int_{0}^{1} \frac{1}{x ^{2} - 2 x - 3} d x

- \frac{1}{4} ln (3)

Decimal Notation

- 0.27465 \dots

Solution steps

\int_{0}^{1} \frac{1}{x ^{2} - 2 x - 3} d x

Take the partial fraction of

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

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

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

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

= - \frac{1}{4} ln (2) + \frac{1}{4} (ln (2) - ln (3))

Simplify

- \frac{1}{4} ln (2) + \frac{1}{4} (ln (2) - ln (3)) : - \frac{1}{4} ln (3)

= - \frac{1}{4} ln (3)

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What is the integral from 0 to 1 of 1/(x^2-2x-3) ?
The integral from 0 to 1 of 1/(x^2-2x-3) is -1/4 ln(3)