What is the integral from 2 to 7 of 1/(x^2-1) ?

The integral from 2 to 7 of 1/(x^2-1) is -(ln(4/3)-ln(3))/2

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integral from 2 to 7 of 1/(x^2-1)

\int_{2}^{7} \frac{1}{x ^{2} - 1} d x

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

Decimal Notation

0.40546 \dots

Solution steps

\int_{2}^{7} \frac{1}{x ^{2} - 1} d x

Factor

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

= \int_{2}^{7} \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}^{7} \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}^{7}

Simplify

- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{2}^{7} : - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{2}^{7}

= - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{2}^{7}

Compute the boundaries:

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

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

Simplify

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

y = \frac{6 x}{x + 2}

\frac{d}{d x} ((1 + 5 x) (1 - 7 x))

\frac{d}{d x} ((lo g_{e} (x))^{x})

y^{^{''}} + 7 y^{^{'}} + 10 y = 0, y (0) = 1, y^{^{'}} (0) = 0

What is the integral from 2 to 7 of 1/(x^2-1) ?
The integral from 2 to 7 of 1/(x^2-1) is -(ln(4/3)-ln(3))/2