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

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

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

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

diverges

Solution steps

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

Apply u-substitution

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{- 1}^{0}

Compute the boundaries:

diverges

= diverges

y^{^{'}} = \frac{( x y + 3 x - y - 3 )}{( x y - 2 x + 4 y - 8 )}

in v erse l a pl a ce \frac{4}{( 3 s - 1 ) ^{5}}

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

\frac{d}{d x} (\frac{4 x + 2}{( 3 - 2 x ) ^{3}})

\frac{\partial}{\partial x} (x e^{14 x y})

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