What is the integral from 0 to 5 of (x-1)/(x+1) ?

The integral from 0 to 5 of (x-1)/(x+1) is -2ln(6)+5

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integral from 0 to 5 of (x-1)/(x+1)

\int_{0}^{5} \frac{x - 1}{x + 1} d x

- 2 ln (6) + 5

Decimal Notation

1.41648 \dots

Solution steps

\int_{0}^{5} \frac{x - 1}{x + 1} d x

\frac{x - 1}{x + 1} = - \frac{2}{x + 1} + 1

= \int_{0}^{5} - \frac{2}{x + 1} + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{0}^{5} \frac{2}{x + 1} d x + \int_{0}^{5} 1 d x

\int_{0}^{5} \frac{2}{x + 1} d x = 2 ln (6)

\int_{0}^{5} 1 d x = 5

= - 2 ln (6) + 5

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What is the integral from 0 to 5 of (x-1)/(x+1) ?
The integral from 0 to 5 of (x-1)/(x+1) is -2ln(6)+5