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

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

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

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

- x - 1 + 2 ln ∣ x + 1 ∣ + C

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

\int 1 d u = u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

= - u + 2 ln ∣ u ∣

Substitute back

u = x + 1

= - (x + 1) + 2 ln ∣ x + 1 ∣

Simplify

= - x - 1 + 2 ln ∣ x + 1 ∣

Add a constant to the solution

= - x - 1 + 2 ln ∣ x + 1 ∣ + C

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