Find the arc length of the curve y = ln x between

Find the arc length of the curve $y = \ln x$ between the points with ordinates $y = \ln\sqrt3$ and $y = \ln\sqrt8$.

We first translate the ordinate bounds into $x$-bounds, then integrate.

Step 1: Find the corresponding $x$-values

$y=\ln\sqrt3 \implies x=\sqrt3$, and $y=\ln\sqrt8 \implies x=\sqrt8=2\sqrt2$.

Step 2: Differentiate

\[y'=\frac1x\]

Step 3: Set up the integral

\[L=\int_{\sqrt3}^{2\sqrt2}\sqrt{1+\frac1{x^2}}\,dx = \int_{\sqrt3}^{2\sqrt2}\frac{\sqrt{x^2+1}}{x}\,dx\]

Step 4: Antiderivative

Using the standard formula $\displaystyle\int\frac{\sqrt{x^2+1}}{x}\,dx=\sqrt{x^2+1}-\ln\left|\frac{1+\sqrt{x^2+1}}{x}\right|+C$:

At $x=2\sqrt2$: $\sqrt{x^2+1}=\sqrt9=3$, giving $3-\ln\dfrac{4}{2\sqrt2}=3-\ln\sqrt2=3-\tfrac12\ln2$.

At $x=\sqrt3$: $\sqrt{x^2+1}=\sqrt4=2$, giving $2-\ln\dfrac{3}{\sqrt3}=2-\ln\sqrt3=2-\tfrac12\ln3$.

Step 5: Combine

\[L=\left(3-\tfrac12\ln2\right)-\left(2-\tfrac12\ln3\right)=1+\tfrac12(\ln3-\ln2)=1+\frac12\ln\frac32\]

\[\boxed{L=1+\frac12\ln\frac32\approx1.203}\]