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

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

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

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

diverges

Solution steps

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

Find undefined (singularity) points:

x = 1, x = - 1

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int \frac{18}{x} d x

f (x) = x^{2} - 2 x, at x = 3

\frac{\partial}{\partial y} (e^{x} + y e^{x y})

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

\frac{d y}{- y ln ( \frac{y}{k} )} = c d x

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