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

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

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

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

diverges

Solution steps

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

Find undefined (singularity) points:

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{\partial}{\partial x} (3 x^{2} y + 2 y)

x = - 5, x = 1, y = 2 x^{2} + 12, y = 0

\int_{1}^{2} x e^{x} d x

f (x) = 5^{s i n (x)}

\int_{1}^{e} ln (11 x) d x

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