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 Notation

- 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

x^{\frac{2}{3}}

\frac{d}{d x} ((x^{3} + 2 x)^{37})

\frac{\partial}{\partial y} (x^{3} y^{4} + x^{4} y^{3})

\frac{d}{d x} (- \frac{10 x}{( x ^{2} - 9 ) ^{2}})

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