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

The integral from 3 to 8 of 1/(x^2-1) is -(2ln(3)-ln(14))/2

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integral from 3 to 8 of 1/(x^2-1)

\int_{3}^{8} \frac{1}{x ^{2} - 1} d x

- \frac{2 ln ( 3 ) - ln ( 14 )}{2}

Decimal Notation

0.22091 \dots

Solution steps

\int_{3}^{8} \frac{1}{x ^{2} - 1} d x

Factor

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

= \int_{3}^{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_{3}^{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}]_{3}^{8}

Simplify

- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{3}^{8} : - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{3}^{8}

= - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{3}^{8}

Compute the boundaries:

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

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

Simplify

= - \frac{2 ln ( 3 ) - ln ( 14 )}{2}

y = x^{2} - 6

x \to \infty lim (0 + x)

\frac{d}{d c} (\frac{sin ( 2 c )}{2 - cos ( 2 c )})

\int \frac{3}{1 - 9 x ^{2}} d x

\int (x^{2} - 3 x - 4) d x

What is the integral from 3 to 8 of 1/(x^2-1) ?
The integral from 3 to 8 of 1/(x^2-1) is -(2ln(3)-ln(14))/2