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

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

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

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

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

Decimal Notation

0.43773 \dots

Solution steps

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

Factor

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

= \int_{2}^{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_{2}^{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}]_{2}^{9}

Simplify

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

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

Compute the boundaries:

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

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

Simplify

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

(x y - x) d x + (x y + y) d y = 0

\int x cos (\frac{nπ x}{3}) d x

\int sin (r) d r

y = 9 e^{x} + x

\frac{d y}{d x} = e^{2 x - 3 y}

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