What is the integral from 0 to x of (1-x)/(1+x) ?

The integral from 0 to x of (1-x)/(1+x) is -x+2ln|x+1|

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integral from 0 to x of (1-x)/(1+x)

\int_{0}^{x} \frac{1 - x}{1 + x} d x

- x + 2 ln ∣ x + 1 ∣

Solution steps

\int_{0}^{x} \frac{1 - x}{1 + x} d x

Apply u-substitution

: \int_{1}^{1 + x} \frac{- u + 2}{u} d u

= \int_{1}^{1 + x} \frac{- u + 2}{u} d u

Expand

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

= \int_{1}^{1 + x} - 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}^{1 + x} 1 d u + \int_{1}^{1 + x} \frac{2}{u} d u

\int_{1}^{1 + x} 1 d u = x

\int_{1}^{1 + x} \frac{2}{u} d u = 2 ln ∣ x + 1 ∣

= - x + 2 ln ∣ x + 1 ∣

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What is the integral from 0 to x of (1-x)/(1+x) ?
The integral from 0 to x of (1-x)/(1+x) is -x+2ln|x+1|