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

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

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integral from 0 to 1 of 1/(sqrt(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{2 ( u + 1 )}{u} d u

= \int_{- 1}^{0} \frac{2 ( u + 1 )}{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{- 1}^{0} \frac{u + 1}{u} d u

Expand

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

= 2 \cdot \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

= 2 (\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

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