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

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

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

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

- 1 + 2 ln (2)

Decimal

0.38629 \dots

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

\int_{1}^{2} 1 d u = 1

\int_{1}^{2} \frac{2}{u} d u = 2 ln (2)

= - 1 + 2 ln (2)

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