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 \frac{e ^{- x}}{1} d x

(- 8, - 3), (- 12, - 3)

\frac{\partial}{\partial x} (7 x - y)

2 y \frac{d y}{d x} + 2 = y^{2} + 2 x

\frac{d y}{d x} = \frac{x}{y}

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