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

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

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

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

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

Decimal Notation

0.42364 \dots

Solution steps

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

Factor

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

= \int_{2}^{8} \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}^{8} \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}^{8}

Simplify

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

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

Compute the boundaries:

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

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

\frac{d}{d x} (\frac{1}{2 x + 3})

x^{2} \cdot cos (x)

\frac{d}{d x} (\frac{x ^{3}}{3} - \frac{1}{4 x})

\frac{d}{d x} (\frac{2}{x ^{2} + 7})

\frac{d}{d x} (- 0.5 x)

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