What is the integral of (2u)/(1-u) ?

The integral of (2u)/(1-u) is -2(-1+u+ln|1-u|)+C

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integral of (2u)/(1-u)

\int \frac{2 u}{1 - u} d u

- 2 (- 1 + u + ln ∣ 1 - u ∣) + C

Solution steps

\int \frac{2 u}{1 - u} d u

Take the constant out:

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

= 2 \cdot \int \frac{u}{1 - u} d u

Apply u-substitution

= 2 \cdot \int - \frac{- v + 1}{v} d v

Take the constant out:

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

= 2 (- \int \frac{- v + 1}{v} d v)

Expand

\frac{- v + 1}{v} : - 1 + \frac{1}{v}

= 2 (- \int - 1 + \frac{1}{v} d v)

Apply the Sum Rule:

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

= 2 (- (- \int 1 d v + \int \frac{1}{v} d v))

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= 2 (- (- v + ln ∣ v ∣))

Substitute back

v = 1 - u

= 2 (- (- (1 - u) + ln ∣ 1 - u ∣))

Simplify

= - 2 (- 1 + u + ln ∣ 1 - u ∣)

Add a constant to the solution

= - 2 (- 1 + u + ln ∣ 1 - u ∣) + C

What is the integral of (2u)/(1-u) ?
The integral of (2u)/(1-u) is -2(-1+u+ln|1-u|)+C