<s>[INST] Finding the length of a Curve The following problem is from the book, Calculus and Analytical Geometer by  Thomas and Finney. It is relatively
early in the book, so I would expect the integration to be easy.  

Find the length of the curve:
  $$ 9x^2 = 4y^3$$
  from $(0,0)$ to $\left(2\sqrt{3},3\right)$.  

Answer:
The formal for the length of a curve is:
$$ L = \int_a^b \sqrt{ 1 + {f'(x)}^2 } \, dx $$
In this case, we have:
\begin{align*}
a &= 0 \\
b &= 2\sqrt{3} \\
y^3 &= \frac{9x^2}{4} \\
f(x) &=  \left( \frac{9x^2}{4} \right) ^ {\frac{1}{3}} \\
f'(x) &= \frac{1}{3} \left( \frac{18x}{4} \right) \left( \frac{9x^2}{4} \right) ^ {-\frac{2}{3}} \\
f'(x) &= \left( \frac{3x}{2} \right) \left( \frac{9x^2}{4} \right) ^ {-\frac{2}{3}} \\
\end{align*}
\begin{align*}
L &= \int_0^{2\sqrt{3}} \sqrt{ 1 +  \left( \frac{9x^2}{4} \right) \left( \frac{9x^2}{4} \right) ^ {-\frac{4}{3}}  } \, dx \\
L &= \int_0^{2\sqrt{3}} \sqrt{ 1 +  \left( \frac{9x^2}{4} \right) ^ {-\frac{1}{3}}  } \, dx \\
L &= \int_0^{2\sqrt{3}} \sqrt{ 1 +  \left( \frac{4}{9x^2} \right) ^ {\frac{1}{3}}  } \, dx \\
\end{align*}
The book's answer is $$ \frac{14}{3} $$
Using an online integral calculator, my integral did not match. What did I do wrong?
I used the following website to do the integration:
https://www.integral-calculator.com/
Their answer is:
$$ \left( \frac{4^\frac{1}{3}}{9^\frac{1}{3}x^{\frac{2}{3}}} + 1 \right) ^ \frac{3}{2} x + C $$
 [/INST] From $9x^2=4y^3$, you can also get $x=\frac23y^{3/2}$. And, if $g(x)=\frac23x^{3/2}$, then\begin{align}\int\sqrt{1+\bigl(g'(x)\bigr)^2}\,\mathrm dx&=\int\sqrt{1+x}\,\mathrm dx\\&=\frac23(1+x)^{3/2}.\end{align}So, the length of the curve is$$\frac23(1+3)^{3/2}-\frac23(1+0)^{3/2}=\frac{16}3-\frac23=\frac{14}3.$$
</s>