What is the integral from 4 to 8 of 1/(x^2-1) ?

The integral from 4 to 8 of 1/(x^2-1) is -(3ln(3)-ln(35))/2

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integral from 4 to 8 of 1/(x^2-1)

\int_{4}^{8} \frac{1}{x ^{2} - 1} d x

- \frac{3 ln ( 3 ) - ln ( 35 )}{2}

Decimal

0.12975 \dots

Solution steps

\int_{4}^{8} \frac{1}{x ^{2} - 1} d x

Factor

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

= \int_{4}^{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_{4}^{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}]_{4}^{8}

Simplify

- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{4}^{8} : - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{4}^{8}

= - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{4}^{8}

Compute the boundaries:

\frac{3 ln ( 3 ) - ln ( 7 ) - ln ( 5 )}{2}

= - \frac{3 ln ( 3 ) - ln ( 7 ) - ln ( 5 )}{2}

Simplify

= - \frac{3 ln ( 3 ) - ln ( 35 )}{2}

- \frac{2}{( 1 - x ) ^{3}}

\int \frac{x ^{3}}{( x ^{2} + 5 ) ^{2}} d x

\int cos^{3} (\frac{x}{15}) d x

\int \frac{10}{( x - 1 ) ( x ^{2} + 9 )} d x

cos (π) (x), 0

What is the integral from 4 to 8 of 1/(x^2-1) ?
The integral from 4 to 8 of 1/(x^2-1) is -(3ln(3)-ln(35))/2