# What is the net area between f(x)=(x^2-5)/(x^3-5x in x in[1,2]  and the x-axis?

Area $= \ln 2 = 0.693147 \text{ }$square unit

#### Explanation:

Given $y = \frac{{x}^{2} - 5}{{x}^{3} - 5 x}$ from $x = 1$ to $x = 2$ and the x-axis

$y = \frac{{x}^{2} - 5}{{x}^{3} - 5 x} = \frac{1}{x}$

Area ${\int}_{1}^{2} \left(\frac{1}{x}\right) \mathrm{dx} = {\left[\ln x\right]}_{1}^{2} = \ln 2 - \ln 1 = \ln 2$

Area $= \ln 2 = 0.693147 \text{ }$square unit

graph{(y-(x^2-5)/(x^3-5x))(y+10000x-10000)(y+10000x-20000)(y-0)=0[-0.1,4,-2,2]}

God bless...I hope the explanation is useful