What is the integral from 0 to 8 of x/(x^2-1) ?

The integral from 0 to 8 of x/(x^2-1) is diverges

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integral from 0 to 8 of x/(x^2-1)

\int_{0}^{8} \frac{x}{x ^{2} - 1} d x

diverges

Solution steps

\int_{0}^{8} \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}^{8} \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

\frac{\partial}{\partial y} (x^{4} + y^{4} - 4 x y + 2)

x + \frac{5}{x}

\int_{0}^{1} \frac{10}{4 y - 1} d y

x \to - 1 lim (ln (x))

\frac{\partial}{\partial x} (4 x^{0.5} y^{0.5})

What is the integral from 0 to 8 of x/(x^2-1) ?
The integral from 0 to 8 of x/(x^2-1) is diverges