Find the arc length of the curve $x = 1/6*(y^2+ 3)^{3/2}$ from $y = 0$ to $y = 1$ I am trying to find the arclength of the curve $$x = 1/6\cdot\left(y^2 + 3\right)^{3/2},\;\; 0\leq y\leq 1$$   I got this far and now I am stuck and don't know what to do next. Any help please?
$$\begin{align} dx & =  \left(1/6\right)\cdot\left(3/2\right)\cdot\left(y^2 + 3\right)^{1/2}\cdot\left(2y\right)\\ \\ 
& =  (1/4)(2y)\cdot\left(y^2 + 3\right)^{1/2}\\ \\
&  =  (y/2)\cdot\left(y^{2}+3\right)^{1/2}\end{align}$$

$$\begin{align}L & = \int_0^1 \sqrt{1 + \left(y/2\left(y^2 + 3\right)^{1/2}\right)^{2}}\\ \\
 &= \int_0^1 \sqrt{1 + \left(y^{2}/4\right)\left(y^2 + 3\right)}\\ \\
 &= \int_0^1 \sqrt{1 + \left(y^{4} + 3y^{2}\right)/4}\end{align}$$ 

Simpson's Rule with n=2 already gives a result correct to four decimal places.
$$\frac16\left(1+4\sqrt{1+\left(\frac{1}{2^4}+3\left(\frac{1}{2^2}\right)\right)/4}+\sqrt{2}\right)\approx1.1336\ldots$$
(A CAS says that the integral is $1.13359\ldots$)
