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

The integral from 0 to 6 of (x-1)/(x+1) is -2ln(7)+6

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integral from 0 to 6 of (x-1)/(x+1)

\int_{0}^{6} \frac{x - 1}{x + 1} d x

- 2 ln (7) + 6

Decimal

2.10817 \dots

Solution steps

\int_{0}^{6} \frac{x - 1}{x + 1} d x

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

= \int_{0}^{6} - \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}^{6} \frac{2}{x + 1} d x + \int_{0}^{6} 1 d x

\int_{0}^{6} \frac{2}{x + 1} d x = 2 ln (7)

\int_{0}^{6} 1 d x = 6

= - 2 ln (7) + 6

\int cos (x) - sin (2 x) d x

\frac{d}{d x} (2 a x - \frac{1}{x ^{2}})

\int \frac{5 x}{x ^{2}} d x

x^{3} - 4 x^{2} + 3 x, 0

(y + x) y^{^{'}} = y

What is the integral from 0 to 6 of (x-1)/(x+1) ?
The integral from 0 to 6 of (x-1)/(x+1) is -2ln(7)+6