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

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

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

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

diverges

Solution steps

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

Apply u-substitution

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

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{0}^{1})

Simplify

= [ln ∣ u ∣]_{0}^{1}

Compute the boundaries:

diverges

= diverges

P (x) = x^{3} - 2 x^{2} + 2 x - 1

\int_{0}^{1} x x^{2} + 8 d x

y = 3 x + 1

3 x sin (x)

\int x^{3} - x d x

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