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

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

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

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

diverges

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{2} (x^{3} + x^{2} + x + 1) d x

\int_{1}^{\infty} e^{- 5 n}

\int_{0.5}^{1.5} 0.5 x d x

\int_{1}^{3} (3 x^{2} - 5 x + 2) d x

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

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