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

Decimal

- 0.38629 \dots

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 9 x^{2} e^{x} d x

x^{2} = e^{x - y}

\frac{d}{d x} (cos (x^{7}))

\frac{d}{d x} (sin^{3} (x^{4}))

2^{x + 2}

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