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Popular Calculus >

integral of u/(1-u)

Solution

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

Solution

1 - u - ln ∣ 1 - u ∣ + C

Solution steps

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

Apply u-substitution

= \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

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

Expand

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

= - \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

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

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= - (- v + ln ∣ v ∣)

Substitute back

v = 1 - u

= - (- (1 - u) + ln ∣ 1 - u ∣)

Simplify

= 1 - u - ln ∣ 1 - u ∣

Add a constant to the solution

= 1 - u - ln ∣ 1 - u ∣ + C

Graph

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Frequently Asked Questions (FAQ)

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