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

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

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

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

diverges

Solution steps

\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}

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

Simplify

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

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

Compute the boundaries:

diverges

= diverges

x \to 0 lim (\frac{3}{e ^{x} + 1})

\frac{\partial}{\partial x} (6 x^{2} + 7 y)

\frac{d}{d x} (∣ x ∣^{3} - x^{2})

x \to \infty lim (\frac{12 x ^{3} - 4 x + 1}{1 + 3 x + 2 x ^{3}})

\frac{8 x - 7}{5 x + 12}

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