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

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

Popular >

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

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

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{1} - \frac{1}{u} d u

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

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{1}^{2})

Simplify

= [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= ln (2)

\int_{1}^{l n (x)} (t^{2} + 3) d t

\int_{- \infty}^{0} x e^{- 3 x} d x

\int_{0}^{2} x e^{4 - x^{2}} d x

\int_{s e c (θ)}^{2} 3 r^{2} sin (θ) d r

\int_{1}^{2} \frac{1}{x} + \frac{1}{x ^{3}} d x

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