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

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

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integral of x/((x-1))

integral

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

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

\frac{- 1}{x + 1}

2 x^{5}

x \cdot x^{2}

\frac{u ^{3}}{( u + 1 )}

(cos (x))^{2}