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

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

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

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

diverges

Solution steps

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

Factor

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

= \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

= - \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}

= - [\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

\int_{5}^{7} x^{2} + x + 1 - 15 d x

\int_{- 8}^{8} ∣ 7 x ∣ - (x^{2} - 8) d x

\int_{2}^{4} 5 x^{4} + 8 x^{3} d x

\int_{- 1}^{1} (3 x - 3) d x

\int_{0}^{1} (2 x e^{x^{2}}) d x

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