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

The integral from 2 to 6 of 1/(x^2-1) is -(ln(7)-ln(15))/2

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

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

- \frac{ln ( 7 ) - ln ( 15 )}{2}

Decimal Notation

0.38107 \dots

Solution steps

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

Factor

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

= \int_{2}^{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_{2}^{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}]_{2}^{6}

Simplify

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

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

Compute the boundaries:

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

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

Simplify

= - \frac{ln ( 7 ) - ln ( 15 )}{2}

\frac{\partial}{\partial y} (e^{y^{2}})

t \cdot \frac{d y}{d t} = t^{3} + 25 t^{3} y

\frac{d}{d x} (\frac{e ^{x^{2}}}{x})

y = \frac{2}{3 x ^{3}}

\frac{d y}{d t} = 0.0572 (400 - y)

What is the integral from 2 to 6 of 1/(x^2-1) ?
The integral from 2 to 6 of 1/(x^2-1) is -(ln(7)-ln(15))/2