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

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

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

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

diverges

Solution steps

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

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{- 1}^{0})

Simplify

= [ln ∣ u ∣]_{- 1}^{0}

Compute the boundaries:

diverges

= diverges

\int_{0}^{π} 1 - sin^{3} (x) d x

\int_{0}^{1} x^{4} (1 + 2 x^{5})^{3} d x

\int_{0}^{2} x^{2} - x d x

\int_{e}^{e^{3}} \frac{1}{x ln ( x )} d x

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

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