What is the integral from-2 to 0 of 1/(1-x) ?

The integral from-2 to 0 of 1/(1-x) is ln(3)

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integral from-2 to 0 of 1/(1-x)

\int_{- 2}^{0} \frac{1}{1 - x} d x

ln (3)

Decimal

1.09861 \dots

Solution steps

\int_{- 2}^{0} \frac{1}{1 - x} d x

Apply u-substitution

: \int_{3}^{1} - \frac{1}{u} d u

= \int_{3}^{1} - \frac{1}{u} d u

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

= - \int_{1}^{3} - \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}^{3} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{1}^{3})

Simplify

= [ln ∣ u ∣]_{1}^{3}

Compute the boundaries:

ln (3)

= ln (3)

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What is the integral from-2 to 0 of 1/(1-x) ?
The integral from-2 to 0 of 1/(1-x) is ln(3)