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

The integral from 7 to 8 of 1/(x^2-1) is -(ln(27/28))/2

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integral from 7 to 8 of 1/(x^2-1)

\int_{7}^{8} \frac{1}{x ^{2} - 1} d x

- \frac{ln ( \frac{27}{28} )}{2}

Decimal Notation

0.01818 \dots

Solution steps

\int_{7}^{8} \frac{1}{x ^{2} - 1} d x

Factor

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

= \int_{7}^{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_{7}^{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}]_{7}^{8}

Simplify

- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{7}^{8} : - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{7}^{8}

= - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{7}^{8}

Compute the boundaries:

\frac{2 ln ( 3 ) - ln ( 7 ) - 3 ln ( 2 ) + ln ( 6 )}{2}

= - \frac{2 ln ( 3 ) - ln ( 7 ) - 3 ln ( 2 ) + ln ( 6 )}{2}

Simplify

= - \frac{ln ( \frac{27}{28} )}{2}

\int \frac{9}{x ( x - 9 ) ^{2}} d x

\int (tan^{2} (x) 4) d x

\frac{d}{d x} (e^{2 x^{6}} - 1)

\frac{\partial}{\partial y} (ln (x^{6} + y^{7}))

x \to 6 - lim (cos (\frac{1}{x - 6}))

What is the integral from 7 to 8 of 1/(x^2-1) ?
The integral from 7 to 8 of 1/(x^2-1) is -(ln(27/28))/2