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

The integral from 5 to 6 of 1/(x^2-1) is -(ln(14/15))/2

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integral from 5 to 6 of 1/(x^2-1)

\int_{5}^{6} \frac{1}{x ^{2} - 1} d x

- \frac{ln ( \frac{14}{15} )}{2}

Decimal Notation

0.03449 \dots

Solution steps

\int_{5}^{6} \frac{1}{x ^{2} - 1} d x

Factor

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

= \int_{5}^{6} \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}^{6} \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}^{6}

Simplify

- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{5}^{6} : - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{5}^{6}

= - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{5}^{6}

Compute the boundaries:

\frac{ln ( 7 ) - ln ( 5 ) - ln ( 6 ) + 2 ln ( 2 )}{2}

= - \frac{ln ( 7 ) - ln ( 5 ) - ln ( 6 ) + 2 ln ( 2 )}{2}

Simplify

= - \frac{ln ( \frac{14}{15} )}{2}

\int_{- 2}^{3} \frac{18}{x ^{4}} d x

x^{2} + 7

\frac{\partial}{\partial x} (3 x^{2} + 5 y^{3})

y^{^{'}} + \frac{y}{t} = 3 cos (4 t)

y^{^{'}} + y + y^{2} = 0

What is the integral from 5 to 6 of 1/(x^2-1) ?
The integral from 5 to 6 of 1/(x^2-1) is -(ln(14/15))/2