What is the integral from 0 to 4 of 1/(x^2-4x-5) ?

The integral from 0 to 4 of 1/(x^2-4x-5) is -1/3 ln(5)

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integral from 0 to 4 of 1/(x^2-4x-5)

\int_{0}^{4} \frac{1}{x ^{2} - 4 x - 5} d x

- \frac{1}{3} ln (5)

Decimal

- 0.53647 \dots

Solution steps

\int_{0}^{4} \frac{1}{x ^{2} - 4 x - 5} d x

Take the partial fraction of

\frac{1}{x ^{2} - 4 x - 5} : - \frac{1}{6 ( x + 1 )} + \frac{1}{6 ( x - 5 )}

= \int_{0}^{4} - \frac{1}{6 ( x + 1 )} + \frac{1}{6 ( x - 5 )} 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}^{4} \frac{1}{6 ( x + 1 )} d x + \int_{0}^{4} \frac{1}{6 ( x - 5 )} d x

\int_{0}^{4} \frac{1}{6 ( x + 1 )} d x = \frac{1}{6} ln (5)

\int_{0}^{4} \frac{1}{6 ( x - 5 )} d x = - \frac{1}{6} ln (5)

= - \frac{1}{6} ln (5) - \frac{1}{6} ln (5)

Simplify

- \frac{1}{6} ln (5) - \frac{1}{6} ln (5) : - \frac{1}{3} ln (5)

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

\frac{d}{d x} (8 x^{\frac{2}{3}} + x^{\frac{5}{3}})

x \to - 4 lim (\frac{1}{4} + \frac{1}{x})

\frac{d}{d x} ((cos (6 x))^{x})

x \to 0 lim (x^{3} ln (x))

x \to 0 lim (\frac{1 - 6 x}{x})

What is the integral from 0 to 4 of 1/(x^2-4x-5) ?
The integral from 0 to 4 of 1/(x^2-4x-5) is -1/3 ln(5)