What is the integral from 4 to 5 of 1/(x^2-1) ?

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

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integral from 4 to 5 of 1/(x^2-1)

\int_{4}^{5} \frac{1}{x ^{2} - 1} d x

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

Decimal Notation

0.05268 \dots

Solution steps

\int_{4}^{5} \frac{1}{x ^{2} - 1} d x

Factor

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

= \int_{4}^{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_{4}^{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}]_{4}^{5}

Simplify

- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{4}^{5} : - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{4}^{5}

= - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{4}^{5}

Compute the boundaries:

\frac{ln ( 6 ) - 2 ln ( 2 ) - ln ( 5 ) + ln ( 3 )}{2}

= - \frac{ln ( 6 ) - 2 ln ( 2 ) - ln ( 5 ) + ln ( 3 )}{2}

Simplify

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

\frac{\partial}{\partial x} (x \cdot ln (x))

\int 2 π y (8 - y^{\frac{3}{2}}) d y

f (x) = (ln (x))^{3}, at x = 11

f (x) = x^{3} + 2 x, at x = 1

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

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