What is this length of this curve using a calculator? I have this question:

Set up an integral that represents the length of the curve.Use a calculator to find the length correct to 4 places.

$$y^2 = lnx$$ and $-1 \leq y \leq 1$
so implicitly differentiating:
$$2y \frac{dy}{dx} = \frac{1}{x}$$
$$\frac{dy}{dx} = \frac{1}{x2y}$$
and $y = \sqrt{lnx}$
So the curve is
$$\int_{-1}^1 \sqrt{1+(\frac{1}{2x\ln{x}})^2}$$
$$\int_{-1}^1 \sqrt{1+(\frac{1}{4x^2\ln{x}^2})}$$
Is that right? I can just plug that into a CAD right?
Wolfram exceeded the time allotted... did I set this up incorrectly?
 A: The limits of integration in the OP are not correct. In fact, inasmuch as $\log(x)$ is not defined for $x\le 0$, the integral in the OP is also not defined.  
To proceed correctly, we note that $y=\sqrt{\log(x)}$ for $y\in[0,1]$ ($x\in [1,e]$) and $y=-\sqrt{\log(x)}$ for $y\in [-1,0]$ ($x\in [1,e]$).  Then, we see that 
$$\begin{align}
\text{Length of Curve}&=2\int_{1}^e \sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\\\\
&=2\int_{1}^e \sqrt{1+\frac{1}{4x^2\log(x)}}\,dx\\\\
&\approx. 4.25523282937328 
\end{align}$$

Alternatively, we have $x=e^{y^2}$ so the 
$$\begin{align}
\text{Length of Curve}&=\int_{-1}^1 \sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy\\\\
&=2\int_0^1 \sqrt{1+4y^2e^{2y^2}}\,dy\\\\
& \approx 4.25523282937328
\end{align}$$
A: What would make this easier is to integrate with respect to $y$, and not $x$. You can re-write this as $x=e^{y^2}$. Now $\frac{dx}{dy}=e^{y^2}.2y$. Hence, the length of the curve can be written as $$\int_{-1}^1 \sqrt{4y^2e^{2y^2}+1}dy$$ You can now plug this into the calculator.
