What is the integral from 0 to 2 of ((x+2))/(x-4) ?

The integral from 0 to 2 of ((x+2))/(x-4) is -6ln(2)+2

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integral from 0 to 2 of ((x+2))/(x-4)

\int_{0}^{2} \frac{( x + 2 )}{x - 4} d x

- 6 ln (2) + 2

Decimal

- 2.15888 \dots

Solution steps

\int_{0}^{2} \frac{x + 2}{x - 4} d x

\frac{x + 2}{x - 4} = \frac{6}{x - 4} + 1

= \int_{0}^{2} \frac{6}{x - 4} + 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{6}{x - 4} d x + \int_{0}^{2} 1 d x

\int_{0}^{2} \frac{6}{x - 4} d x = - 6 ln (2)

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

= - 6 ln (2) + 2

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What is the integral from 0 to 2 of ((x+2))/(x-4) ?
The integral from 0 to 2 of ((x+2))/(x-4) is -6ln(2)+2