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

\frac{\partial}{\partial x} (36 - 6 x^{2} - y^{2})

\frac{2 x}{e ^{2 x^{2}} - 1}

\frac{d}{d x} (23 sin (x) + 2 cos (x))

\int_{1}^{2} 3 d x

(x^{2} + y^{2})^{2} = 4 x^{2} y, (- 1, 1)

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