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

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

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integral from 1 to 2 of x/(x-1)

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

diverges

Solution steps

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

Apply u-substitution

: \int_{0}^{1} \frac{u + 1}{u} d u

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

Expand

\frac{u + 1}{u} : 1 + \frac{1}{u}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{7 - x e ^{x}}{x + e ^{x}}

x \to 1 lim (\frac{x - 1}{x - 1})

\int (x y^{2} + x^{2} + y + 4) d x

(\frac{x}{x ^{5} + 1})^{^{'}}

f (v) = (v^{- 3} + 3 v^{- 2})^{3}

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