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

The integral from 0 to 1 of 2/(1-x^2) is diverges

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

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

diverges

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{1} \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}

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

Simplify

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

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

Compute the boundaries:

diverges

= diverges

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