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

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

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

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

- 2 ln (2) + 1

Solution steps

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

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

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

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

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

= - 2 ln (2) + 1

\int_{1}^{4} (5 - 2 t + 3 t^{2}) d t

\int 3 x^{2} cos (x^{3}) d x

f (x) = 5 e^{4 x}

x \frac{d y}{d x} + 2 y = - 3 e^{x}

\frac{d}{d x} (7 - 4 x)

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