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

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

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

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

\frac{ln ( 2 )}{2}

Decimal Notation

0.34657 \dots

Solution steps

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

Factor

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

= \int_{2}^{5} \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}^{5} \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}^{5}

Simplify

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

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

Compute the boundaries:

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

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

Simplify

= \frac{ln ( 2 )}{2}

\frac{\partial}{\partial z} (e^{2 z} x + 2 z)

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

x \to + 0 - lim (e^{\frac{7}{x}})

\frac{\partial}{\partial x} (- \frac{1}{( 10 + x \cdot y )})

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