What is the integral from 8 to 9 of 1/(x^2-1) ?

The integral from 8 to 9 of 1/(x^2-1) is -(ln(5/4)-2ln(3)+ln(7))/2

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integral from 8 to 9 of 1/(x^2-1)

\int_{8}^{9} \frac{1}{x ^{2} - 1} d x

- \frac{ln ( \frac{5}{4} ) - 2 ln ( 3 ) + ln ( 7 )}{2}

Decimal Notation

0.01408 \dots

Solution steps

\int_{8}^{9} \frac{1}{x ^{2} - 1} d x

Factor

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

= \int_{8}^{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_{8}^{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}]_{8}^{9}

Simplify

- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{8}^{9} : - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{8}^{9}

= - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{8}^{9}

Compute the boundaries:

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

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

Simplify

= - \frac{ln ( \frac{5}{4} ) - 2 ln ( 3 ) + ln ( 7 )}{2}

y = 9 - x^{2}, 0 \leq x \leq 2

f (x) = 7 e^{x^{2} - 9 x + 8}, at x = 0

y = 164 4 x^{4} + 4

y^{^{''}} + 6 y^{^{'}} + 5 y = 10

\int \frac{1}{h} d h

What is the integral from 8 to 9 of 1/(x^2-1) ?
The integral from 8 to 9 of 1/(x^2-1) is -(ln(5/4)-2ln(3)+ln(7))/2