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

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

Popular >

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

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

diverges

Solution steps

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

Apply u-substitution

: \int_{2}^{0} - \frac{1}{u} d u

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

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{0}^{2})

Simplify

= [ln ∣ u ∣]_{0}^{2}

Compute the boundaries:

diverges

= diverges

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

\int_{0}^{\frac{π}{2}} 3 sin (2 x) d x

\int_{0}^{a} sin^{2} (\frac{2 π x}{a}) d x

\int_{0}^{a} sin^{2} (x) d x

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

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