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

The integral from 2 to 5/2 of 1/(1-x^2) is (ln(7/2)-ln(9/2))/2

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

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

\frac{ln ( \frac{7}{2} ) - ln ( \frac{9}{2} )}{2}

Decimal

- 0.12565 \dots

Solution steps

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

Use the common integral:

\int \frac{1}{1 - x ^{2}} d x = \frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}

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

Simplify

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

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

Compute the boundaries:

\frac{ln ( \frac{7}{2} ) - ln ( \frac{3}{2} ) - ln ( 3 )}{2}

= \frac{ln ( \frac{7}{2} ) - ln ( \frac{3}{2} ) - ln ( 3 )}{2}

Simplify

= \frac{ln ( \frac{7}{2} ) - ln ( \frac{9}{2} )}{2}

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What is the integral from 2 to 5/2 of 1/(1-x^2) ?
The integral from 2 to 5/2 of 1/(1-x^2) is (ln(7/2)-ln(9/2))/2