What is the integral from 0 to 1/4 of 1/(1-x^2) ?

The integral from 0 to 1/4 of 1/(1-x^2) is (ln(5/4)-ln(3/4))/2

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integral from 0 to 1/4 of 1/(1-x^2)

\int_{0}^{\frac{1}{4}} \frac{1}{1 - x ^{2}} d x

\frac{ln ( \frac{5}{4} ) - ln ( \frac{3}{4} )}{2}

Decimal

0.25541 \dots

Solution steps

\int_{0}^{\frac{1}{4}} \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}]_{0}^{\frac{1}{4}}

Simplify

[\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{0}^{\frac{1}{4}} : [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{0}^{\frac{1}{4}}

= [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{0}^{\frac{1}{4}}

Compute the boundaries:

\frac{ln ( \frac{5}{4} ) - ln ( \frac{3}{4} )}{2}

= \frac{ln ( \frac{5}{4} ) - ln ( \frac{3}{4} )}{2}

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\int_{3}^{5} x - 3 d x

What is the integral from 0 to 1/4 of 1/(1-x^2) ?
The integral from 0 to 1/4 of 1/(1-x^2) is (ln(5/4)-ln(3/4))/2