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

The integral from 0 to 2 of (x-1)/(x+1) is -2ln(3)+2

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

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

- 2 ln (3) + 2

Decimal

- 0.19722 \dots

Solution steps

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

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

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

\int_{0}^{2} \frac{2}{x + 1} d x = 2 ln (3)

\int_{0}^{2} 1 d x = 2

= - 2 ln (3) + 2

ln (\frac{2 x + 3}{3 x - 4})

\int \frac{7}{35 sin ( x ) - 12 cos ( x )} d x

\int_{0}^{t} e^{- x^{2}} d x

n = 0 \sum \infty (- 3 x)^{n}

x \to 0 lim (1 - cos^{2} (x))

What is the integral from 0 to 2 of (x-1)/(x+1) ?
The integral from 0 to 2 of (x-1)/(x+1) is -2ln(3)+2