What is the integral from 0 to 4 of x/(x^2-1) ?

The integral from 0 to 4 of x/(x^2-1) is diverges

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integral from 0 to 4 of x/(x^2-1)

\int_{0}^{4} \frac{x}{x ^{2} - 1} d x

diverges

Solution steps

\int_{0}^{4} \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}^{4} \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 x} (3 x + 2 y - 3)

y = x^{2}, y = 3 - x, (0, 1.3)

f (x) = 3 x^{2} e^{x} + e^{x} (x^{3} + 1)

\int - \frac{8}{x ^{2}} d x

y^{^{'}} = 18.2 - 0.7 y

What is the integral from 0 to 4 of x/(x^2-1) ?
The integral from 0 to 4 of x/(x^2-1) is diverges