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

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

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

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

diverges

Solution steps

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

f (x) = - 2 x^{3}, (- 1, 2)

\frac{9}{( s + 3 ) ^{3}}

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

y^{^{'}} + \frac{1}{x} y = 3 x^{2}

x \to 3 lim (5 x - 1)

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