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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x/(x-1))

\int (\frac{x}{x - 1}) d x

x - 1 + ln ∣ x - 1 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{u + 1}{u} d u

= \int \frac{u + 1}{u} d u

Expand

\frac{u + 1}{u} : 1 + \frac{1}{u}

= \int 1 + \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d u + \int \frac{1}{u} d u

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= u + ln ∣ u ∣

Substitute back

u = x - 1

= x - 1 + ln ∣ x - 1 ∣

Add a constant to the solution

= x - 1 + ln ∣ x - 1 ∣ + C

x^{3} - 25 x + 5, at 5

x \to 0 lim (cos (x) - 1 + x)

cos^{2} (2 t)

\int_{- 2}^{2} (12 - 3 x^{2}) d x

\frac{d}{d x} (3 tan (4 x))