What is the integral from 0 to 0.8 of 1/(1-x^2) ?

The integral from 0 to 0.8 of 1/(1-x^2) is 1.09861…

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integral from 0 to 0.8 of 1/(1-x^2)

\int_{0}^{0.8} \frac{1}{1 - x ^{2}} d x

1.09861 \dots

Solution steps

\int_{0}^{0.8} \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}^{0.8}

Simplify

[\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{0}^{0.8} : [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{0}^{0.8}

= [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{0}^{0.8}

Compute the boundaries:

1.09861 \dots

= 1.09861 \dots

What is the integral from 0 to 0.8 of 1/(1-x^2) ?
The integral from 0 to 0.8 of 1/(1-x^2) is 1.09861…