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

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

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integral from 3 to 5 of 1/(x^2-1)

\int_{3}^{5} \frac{1}{x ^{2} - 1} d x

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

Solution steps

\int_{3}^{5} \frac{1}{x ^{2} - 1} d x

Factor

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

= \int_{3}^{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_{3}^{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}]_{3}^{5}

Simplify

- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{3}^{5} : - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{3}^{5}

= - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{3}^{5}

Compute the boundaries:

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

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

Simplify

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

y = x^{2}, y = - 2 x^{2} + 12

\int_{l n (1)}^{l n (2)} e^{3 x} d x

\frac{d}{d x} (\frac{6 x ^{5}}{5} - 7 x)

x^{2} e^{2 x}

\frac{d}{d x} (2 sec^{2} (2 x))

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