arc length of logarithmic curve

The arc length of the graph of logarithm function (http://planetmath.org/NaturalLogarithm2) is expressible in closed form (other cases are listed in the entry arc length of parabola).  The usual arc length

$$s={\int }_a^b\sqrt{1+{(f^{\prime }(x))}^2}𝑑x$$

gives, if  ,  for  $f(x):=\ln x$,  $f^{\prime }(x)=\frac{1}{x}$,  the expression

$s={\displaystyle {\int }_a^b}{\displaystyle \frac{\sqrt{1+x^2}}{x}}𝑑x.$(1)

Here, finding a suitable substitution for integration may be a bit difficult.  E.g.  $x:=\tan t$leads to

$$\int \frac{\sqrt{1+x^2}}{x}𝑑x=\int \frac{dt}{\sin t{\cos }^{2}t},$$

the substitution  $x:=\sinh t$  to

$$\int \frac{\sqrt{1+x^2}}{x}𝑑x=\int \frac{{\cosh }^{2}t}{\sinh t}𝑑t,$$

which both seem to require a new substitution.  As well the Euler’s substitutions (1st and 2nd ones) lead to awkward rational functions as integrands.

But there is the straightforward substitution

$$\sqrt{1+x^2}:=t,x=\sqrt{t^2-1},dx=\frac{tdt}{\sqrt{t^2-1}}$$

yielding

$$\int \frac{\sqrt{1+x^2}}{x}𝑑x=\int \frac{t^{2}dt}{t^2-1}=t+\frac{1}{2}\ln \frac{t-1}{t+1}+C=t-\mathrm{arcoth}t+C$$

(see area functions) and then

$$\int \frac{\sqrt{1+x^2}}{x}𝑑x=\sqrt{1+x^2}+\frac{1}{2}\ln \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}+C=\sqrt{1+x^2}+\ln \frac{x}{1+\sqrt{1+x^2}}+C.$$

Using this antiderivative, one can obtain the arc length (1).  For example, if  $a=\sqrt{3}$  and $b=\sqrt{15}$,  the result is  $s=2+\ln \frac{3}{\sqrt{5}}$.

As for finding the arc length of the graph of the http://planetmath.org/node/2541exponential function  $x\mapsto e^x$,  which actually is the same curve as the graph of the inverse function  $x\mapsto \ln x$,  one may write the expression

$s={\displaystyle {\int }_{\alpha }^{\beta }}\sqrt{1+e^{2x}}𝑑x.$(2)

Since here the substitution

$$e^x:=t,x=\ln t,dx=\frac{dt}{t}$$

shows that

$$\int \sqrt{1+e^{2x}}𝑑x=\int \frac{\sqrt{1+t^2}}{t}𝑑t,$$

we see that it’s really a question of the same task as above.  The antiderivative is

$$\int \sqrt{1+e^{2x}}𝑑x=\sqrt{1+e^{2x}}-\mathrm{arsinh}{e}^{-x}+C=\sqrt{1+e^{2x}}+\ln \frac{e^x}{1+\sqrt{1+e^{2x}}}+C.$$