What is the integral of 1/(1-u) ?

The integral of 1/(1-u) is -ln|1-u|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(1-u)

\int \frac{1}{1 - u} d u

- ln ∣ 1 - u ∣ + C

Solution steps

\int \frac{1}{1 - u} d u

Apply u-substitution

: \int - \frac{1}{v} d v

= \int - \frac{1}{v} d v

Take the constant out:

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

= - \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - ln ∣ v ∣

Substitute back

v = 1 - u

= - ln ∣ 1 - u ∣

Add a constant to the solution

= - ln ∣ 1 - u ∣ + C

\int e^{- 1 x} d x

49 y^{6} - 84 y^{3} + 36

\int \frac{3}{x ( x ^{4} + 7 )} d x

\int e^{7 x + 8} d x

\int (e^{x} sin (y) - y) d x