What is the integral from 4 to 9 of 1/(x^2-1) ?

The integral from 4 to 9 of 1/(x^2-1) is -(ln(3/4))/2

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integral from 4 to 9 of 1/(x^2-1)

\int_{4}^{9} \frac{1}{x ^{2} - 1} d x

- \frac{ln ( \frac{3}{4} )}{2}

Decimal Notation

0.14384 \dots

Solution steps

\int_{4}^{9} \frac{1}{x ^{2} - 1} d x

Factor

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

= \int_{4}^{9} \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}^{9} \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}^{9}

Simplify

- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{4}^{9} : - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{4}^{9}

= - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{4}^{9}

Compute the boundaries:

\frac{ln ( 10 ) - 3 ln ( 2 ) - ln ( 5 ) + ln ( 3 )}{2}

= - \frac{ln ( 10 ) - 3 ln ( 2 ) - ln ( 5 ) + ln ( 3 )}{2}

Simplify

= - \frac{ln ( \frac{3}{4} )}{2}

\int_{0}^{1} 4 e^{7 x} d x

1 - 6 x^{2} y^{^{'}} = x

x = - 2, y = x + 2, y = x^{2}

\int 8 y d y

m = 2 \sum \infty \frac{5}{6 ^{m}}

What is the integral from 4 to 9 of 1/(x^2-1) ?
The integral from 4 to 9 of 1/(x^2-1) is -(ln(3/4))/2