# What is the arc length of f(x) = 3xln(x^2)  on x in [1,3] ?

Mar 18, 2018

19.88

#### Explanation:

First, we can observe that $\ln \left({x}^{2}\right) = 2 \ln \left(x\right)$, simplifying the above equation a little bit to be $f \left(x\right) = 6 x \ln x$.

For future use, we can calculate its derivative using product rule, getting $f ' \left(x\right) = 6 \left(\ln \left(x\right) + 1\right)$

We now consider an infinitesimal arclength, $\mathrm{ds}$. We know, by Pythagorean theorem,
${\mathrm{ds}}^{2} = {\mathrm{dx}}^{2} + {\mathrm{df}}^{2}$
We can actually factor out a common factor
${\mathrm{ds}}^{2} = {\mathrm{dx}}^{2} + {\mathrm{dx}}^{2} \cdot {\left(\frac{\mathrm{df}}{\mathrm{dx}}\right)}^{2}$
and now
$\mathrm{ds} = \mathrm{dx} \sqrt{1 + {\left(\frac{\mathrm{df}}{\mathrm{dx}}\right)}^{2}}$

Now, we can integrate! Let the total arc length be $S$:
$S = \int \mathrm{ds} = {\int}_{1}^{3} \sqrt{1 + {\left(\frac{\mathrm{df}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$
$S = {\int}_{1}^{3} \sqrt{1 + 36 {\left(\ln x + 1\right)}^{2}} \mathrm{dx}$
$= {\int}_{1}^{3} \sqrt{1 + 36 {\ln}^{2} x + 72 \ln x + 36} \mathrm{dx}$
$= {\int}_{1}^{3} \sqrt{36 {\ln}^{2} x + 72 \ln x + 37} \mathrm{dx}$

Now this integral looks very ugly. And it is! This doesn't have an analytical (exact) answer, as far as I can tell.

We can't prove that this doesn't have a clean answer easily. I proved it to myself by using u-subsitution with u = ln(x) and searching an integration table for integrals of functions of the square root of quadratics times an exponential with no luck.

Anyway, that means we just have to plug this into a calculator to get an answer:

$S \approx 19.88$