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

The integral from 6 to 9 of 1/(x^2-1) is -(ln(5/4)-ln(7)+ln(5))/2

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integral from 6 to 9 of 1/(x^2-1)

\int_{6}^{9} \frac{1}{x ^{2} - 1} d x

- \frac{ln ( \frac{5}{4} ) - ln ( 7 ) + ln ( 5 )}{2}

Decimal

0.05666 \dots

Solution steps

\int_{6}^{9} \frac{1}{x ^{2} - 1} d x

Factor

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

= \int_{6}^{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_{6}^{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}]_{6}^{9}

Simplify

- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{6}^{9} : - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{6}^{9}

= - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{6}^{9}

Compute the boundaries:

\frac{ln ( 10 ) - 3 ln ( 2 ) - ln ( 7 ) + ln ( 5 )}{2}

= - \frac{ln ( 10 ) - 3 ln ( 2 ) - ln ( 7 ) + ln ( 5 )}{2}

Simplify

= - \frac{ln ( \frac{5}{4} ) - ln ( 7 ) + ln ( 5 )}{2}

What is the integral from 6 to 9 of 1/(x^2-1) ?
The integral from 6 to 9 of 1/(x^2-1) is -(ln(5/4)-ln(7)+ln(5))/2