TITLE: Arc length of logarithm function
QUESTION [8 upvotes]: I need to find the length of $y = \ln(x)$ (natural logarithm) from $x=\sqrt3$ to $x=\sqrt8$.
So, if I am not mistake, the length should be $$\int^\sqrt8_\sqrt3\sqrt{1+\frac{1}{x^2}}dx$$
I am having trouble calculating the integral. I tried to do substitution, but I still fail to think of a way to integrate it. This is what I have done so far:
$$\sqrt{1+\frac{1}{x^2}}=u-\frac{1}{x}$$
$$x=\frac{2u}{u-1}$$
$$dx=\frac{2}{(u-1)^2}du$$
$$\sqrt{1+\frac{1}{x^2}}=u-\frac{1}{x}=u-\frac{1}{2}+\frac{1}{2u}$$
$$\int\sqrt{1+\frac{1}{x^2}}dx=2\int\frac{u-\frac{1}{2}+\frac{1}{2u}}{(u-1)^2}du$$
And I am stuck.

REPLY [2 votes]: $\int^{\sqrt{8}}_{\sqrt{3}}\sqrt{1+\frac{1}{x^{2}}}dx$=$\int^{\sqrt{8}}_{\sqrt{3}}\frac{\sqrt{1+x^{2}}}{x}dx$=$\int^{\sqrt{8}}_{\sqrt{3}}\frac{1+x^{2}}{x\sqrt{1+x^{2}}}$=$\int^{\sqrt{8}}_{\sqrt{3}}\frac{x}{\sqrt{1+x^{2}}}dx$+
+$\int^{\sqrt{8}}_{\sqrt{3}}\frac{1}{x\sqrt{1+x^{2}}}dx$=
$=\frac{1}{2}\int^{\sqrt{8}}_{\sqrt{3}}\frac{(1+x^{2})'}{\sqrt{1+x^{2}}}dx$
$-\int^{\sqrt{8}}_{\sqrt{3}}\frac{(\frac{1}{x})'}{\sqrt{1+\frac{1}{x^{2}}}}dx =$ $\sqrt{1+x^{2}}|^{\sqrt{8}}_{\sqrt{3}}-ln(\frac{1}{x}+
\sqrt{1+\frac{1}{x^{2}}})$ $|^{\sqrt{8}}_{\sqrt{3}}$
$=1+\frac{1}{2}$$ln\frac{3}{2}$