What is the integral from 4 to 6 of 1/(x^2-1) ?

The integral from 4 to 6 of 1/(x^2-1) is -(ln(7)-2ln(5)+ln(3))/2

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integral from 4 to 6 of 1/(x^2-1)

\int_{4}^{6} \frac{1}{x ^{2} - 1} d x

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

Decimal Notation

0.08717 \dots

Solution steps

\int_{4}^{6} \frac{1}{x ^{2} - 1} d x

Factor

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

= \int_{4}^{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_{4}^{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}]_{4}^{6}

Simplify

- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{4}^{6} : - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{4}^{6}

= - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{4}^{6}

Compute the boundaries:

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

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

\frac{\partial}{\partial x} (t^{8} e^{- x})

\frac{d}{d x} (\frac{x + 1}{x - 1})

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

\frac{d}{d x} ((1 + sin (x)) \cdot cos (2 x))

\int \frac{2 x ^{2} + 3 x + 7}{x ^{3} + 3 x} d x

What is the integral from 4 to 6 of 1/(x^2-1) ?
The integral from 4 to 6 of 1/(x^2-1) is -(ln(7)-2ln(5)+ln(3))/2