What is the integral from 5 to 9 of 1/(x^2-1) ?

The integral from 5 to 9 of 1/(x^2-1) is -(ln(5/6))/2

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integral from 5 to 9 of 1/(x^2-1)

\int_{5}^{9} \frac{1}{x ^{2} - 1} d x

- \frac{ln ( \frac{5}{6} )}{2}

Decimal

0.09116 \dots

Solution steps

\int_{5}^{9} \frac{1}{x ^{2} - 1} d x

Factor

x^{2} - 1 : - (- x^{2} + 1)

= \int_{5}^{9} \frac{1}{- ( - x ^{2} + 1 )} d x

Take the constant out:

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

= - \int_{5}^{9} \frac{1}{- x ^{2} + 1} d x

Use the common integral:

\int \frac{1}{- x ^{2} + 1} d x = \frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}

= - [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{5}^{9}

Simplify

- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{5}^{9} : - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{5}^{9}

= - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{5}^{9}

Compute the boundaries:

\frac{- ln ( 2 ) - ln ( 6 ) + ln ( 10 )}{2}

= - \frac{- ln ( 2 ) - ln ( 6 ) + ln ( 10 )}{2}

Simplify

= - \frac{ln ( \frac{5}{6} )}{2}

3 2 + 2 x + x^{3}

n = 3 \sum \infty \frac{3}{4} \cdot \frac{1}{4 ^{n}}

\frac{\partial}{\partial z} (x z)

- x^{2} + 3 x + 1, - x + 1

t e^{- t}

What is the integral from 5 to 9 of 1/(x^2-1) ?
The integral from 5 to 9 of 1/(x^2-1) is -(ln(5/6))/2