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

The integral from 3 to 9 of 1/(x^2-1) is -(ln(5/8))/2

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integral from 3 to 9 of 1/(x^2-1)

\int_{3}^{9} \frac{1}{x ^{2} - 1} d x

- \frac{ln ( \frac{5}{8} )}{2}

Decimal Notation

0.23500 \dots

Solution steps

\int_{3}^{9} \frac{1}{x ^{2} - 1} d x

Factor

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

= \int_{3}^{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_{3}^{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}]_{3}^{9}

Simplify

- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{3}^{9} : - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{3}^{9}

= - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{3}^{9}

Compute the boundaries:

\frac{ln ( 10 ) - 4 ln ( 2 )}{2}

= - \frac{ln ( 10 ) - 4 ln ( 2 )}{2}

Simplify

= - \frac{ln ( \frac{5}{8} )}{2}

\frac{\partial}{\partial x} (4 - x^{2} - y^{2})

\int \frac{3 - cos ( x )}{sin ^{2} ( x )} d x

\int_{- 8}^{0} \frac{1}{( x - 8 ) ^{2}} d x

\frac{d y}{d x} = \frac{( y ^{2} + 4 )}{2 x y}

\frac{d y}{d x} = \frac{cos ( 8 x )}{cos ( 4 y )}

What is the integral from 3 to 9 of 1/(x^2-1) ?
The integral from 3 to 9 of 1/(x^2-1) is -(ln(5/8))/2