What is the integral from 5 to 8 of 1/(x^2-1) ?

The integral from 5 to 8 of 1/(x^2-1) is -(ln(6/7))/2

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integral from 5 to 8 of 1/(x^2-1)

\int_{5}^{8} \frac{1}{x ^{2} - 1} d x

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

Decimal Notation

0.07707 \dots

Solution steps

\int_{5}^{8} \frac{1}{x ^{2} - 1} d x

Factor

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

= \int_{5}^{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_{5}^{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}]_{5}^{8}

Simplify

- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{5}^{8} : - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{5}^{8}

= - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{5}^{8}

Compute the boundaries:

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

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

Simplify

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

\frac{d y}{d x} = x y + 5 x

x \to 0 lim (\frac{x}{sin ( 5 x )})

\int_{0}^{4} \frac{1}{x ^{2} - x - 2} d x

\frac{2}{s ( s ^{2} + 4 s + 3 )}

x \to \infty lim (\frac{6 x ^{2} + 2 x}{2 x ^{2} - 2 x + 1})

What is the integral from 5 to 8 of 1/(x^2-1) ?
The integral from 5 to 8 of 1/(x^2-1) is -(ln(6/7))/2