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The arc length of the graph of logarithm function is expressible in closed form (other cases are listed in the entry arc length of parabola). The usual arc length formula $$s \;=\; \int_a^b\!\sqrt{1+(f'(x))^2}\,dx$$ gives, if $0 < a < b$ , for $f(x) := \ln{x}$ , $f'(x) = \frac{1}{x}$ , the expression
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(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 simple substitution $$\sqrt{1\!+\!x^2} \;:=\; t, \quad x \;=\; \sqrt{t^2\!-\!1}, \quad dx \;=\; \frac{t\,dt}{\sqrt{t^2\!-\!1}}$$ yielding $$\int\!\frac{\sqrt{1\!+\!x^2}}{x}\,dx \;=\; \int\!\frac{t^2\,dt}{t^2\!-\!1} \;=\; t+\frac{1}{2}\ln\frac{t\!-\!1}{t\!+\!1}+C \;=\; t-\arcoth{t}+C$$ (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 exponential 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
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(2) |
Since here the substitution $$e^x \;:=\; t, \quad x \;=\; \ln{t}, \quad dx \;=\; \frac{dt}{t}$$ 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 $$\int\!\sqrt{1\!+\!e^{2x}}\,dx \;=\; \sqrt{1\!+\!e^{2x}}-\arsinh{e^{-x}}+C \;=\; \sqrt{1\!+\!e^{2x}}+\ln\frac{e^x}{1+\sqrt{1\!+\!e^{2x}}}+C.$$
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