[Math] Arc length of the curve y=ln(x)

Question

Q: Find the arc length of the curve $y=\ln(x)$ where $x$ ranges from $\sqrt{3}$ to $\sqrt{15}$.

I think I am stuck in calculation part.

The answer is $2 + \ln(3) – \frac{1}{2}\ln(5)$. But I can't derive that from my last line.

help me, please.

Answer 1

Your problem now is how to evaluate the integral $$\int_{\sqrt{3}}^{\sqrt{15}}\frac{\sqrt{x^2+1}}{x}dx.$$

Let $$F(x)=\int\frac{\sqrt{x^2+1}}{x}dx.$$ Let $x=\tan\theta$. All your computations after this substitution are all correct. Note that $$\csc\theta=\frac{\sqrt{x^2+1}}{x}$$ $$\cot\theta=\frac{1}{x}$$ and $$\sec\theta=\sqrt{x^2+1}.$$ Hence, we get $$F(x)=-\ln\left(\frac{\sqrt{x^2+1}+1}{x} \right)+\sqrt{x^2+1}+C.$$ Thus, $$ \begin{align} \int_{\sqrt{3}}^{\sqrt{15}}\frac{\sqrt{x^2+1}}{x}dx&=F(\sqrt{15})-F(\sqrt{3})\\ &=\left[-\ln\left(\frac{5}{\sqrt{15}}\right)+4+C\right]-\left[-\ln\left(\frac{3}{\sqrt{3}}\right)+2+C\right]\\ &=\ln\left(\frac{3}{\sqrt{3}}\right)-\ln\left(\frac{5}{\sqrt{15}}\right) +2\\ &=\ln\left[\frac{3}{\sqrt{3}}\div\frac{5}{\sqrt{15}}\right]+2\\ &=\ln\left[\frac{3}{\sqrt{3}} \cdot \frac{\sqrt{15}}{5}\right]+2\\ &=\ln\left[\frac{3}{1} \cdot \frac{\sqrt{5}}{5}\right]+2\\ &=\ln\left[\frac{3}{1} \cdot \frac{1}{\sqrt{5}}\right]+2\\ &=\ln\left(\frac{3}{\sqrt{5}}\right)+2\\ &=\ln 3-\ln\sqrt{5}+2\\ &=\ln 3-\frac{1}{2}(\ln 5)+2. \end{align}$$