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

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

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

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

diverges

Solution steps

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

Factor

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

= \int_{- 1}^{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_{- 1}^{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}]_{- 1}^{1}

Simplify

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

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

Compute the boundaries:

diverges

= diverges

n = 0 \sum \infty 1800 (\frac{1}{3})^{n}

f (x) = 4 x e^{x}, (0, 0)

\int x (4 + x)^{10} d x

\frac{2 ( 1 - ln ( x ) )}{x ^{2}}

x \to - 2 lim (3 x^{2} + 5 x - 1)

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