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

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

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

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

diverges

Solution steps

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

Factor

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

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

= 2 (- \int_{0}^{1} \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}

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