# What is the arc length of f(x)=(3x)/sqrt(x-1)  on x in [2,6] ?

Jul 8, 2018

$\approx 4.514518184$

#### Explanation:

Writing

$f \left(x\right) = 3 x {\left(x - 1\right)}^{- \frac{1}{2}}$
then by the product and the chain rule we get

$f ' \left(x\right) = 3 {\left(x - 1\right)}^{- \frac{1}{2}} + 3 x \cdot \left(- \frac{1}{2}\right) \cdot {\left(x - 1\right)}^{- \frac{3}{2}}$

which simplifies to

$f ' \left(x\right) = \frac{3 \left(x - 2\right)}{2 \cdot {\left(x - 1\right)}^{\frac{3}{2}}}$
so we have to solve

${\int}_{2}^{6} \sqrt{1 + {\left(3 \frac{x - 2}{2 \cdot {\left(x - 1\right)}^{\frac{3}{2}}}\right)}^{2}} \mathrm{dx}$
by a numerical method we get

$\setminus \approx 4.514518184$