What is the integral from 0 to 3 of x/(x^2-1) ?

The integral from 0 to 3 of x/(x^2-1) is diverges

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integral from 0 to 3 of x/(x^2-1)

\int_{0}^{3} \frac{x}{x ^{2} - 1} d x

diverges

Solution steps

\int_{0}^{3} \frac{x}{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{x}{x ^{2} - 1} d x + \int_{1}^{3} \frac{x}{x ^{2} - 1} d x

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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

\int_{1}^{4} π (\frac{2}{y})^{2} d y

\int_{0}^{1} 1 + r^{2} r d r

\int_{- 1}^{4} (- x^{2} + 6 x + 8 - x^{2}) d x

\int_{0}^{1} x^{- \frac{3}{2}} d x

What is the integral from 0 to 3 of x/(x^2-1) ?
The integral from 0 to 3 of x/(x^2-1) is diverges