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

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

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

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

- 1 - u + 2 ln ∣ 1 + u ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{- v + 2}{v} d v

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

Expand

\frac{- v + 2}{v} : - 1 + \frac{2}{v}

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

\int 1 d v = v

\int \frac{2}{v} d v = 2 ln ∣ v ∣

= - v + 2 ln ∣ v ∣

Substitute back

v = 1 + u

= - (1 + u) + 2 ln ∣ 1 + u ∣

Simplify

= - 1 - u + 2 ln ∣ 1 + u ∣

Add a constant to the solution

= - 1 - u + 2 ln ∣ 1 + u ∣ + C

x \to 0 lim (2 sin (x) - 1)

\int (\frac{ln ( x )}{x ^{2}}) d x

\frac{d}{d z} (1 - z^{2})

\int_{- 1}^{2} \frac{x ^{3}}{3} d x

\int x e^{- 9 x} d x

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