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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.

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  • $\begingroup$ I keep hearing $x=\tan (u)$ in my head. I haven't tried it, though. $\endgroup$ – Git Gud May 12 '13 at 20:05
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$\int^{\sqrt{8}}_{\sqrt{3}}\sqrt{1+\frac{1}{x^{2}}}dx$=$\int^{\sqrt{8}}_{\sqrt{3}}\frac{\sqrt{1+x^{2}}}{x}dx$=$\int^{\sqrt{8}}_{\sqrt{3}}\frac{1+x^{2}}{x\sqrt{1+x^{2}}}$=$\int^{\sqrt{8}}_{\sqrt{3}}\frac{x}{\sqrt{1+x^{2}}}dx$+ +$\int^{\sqrt{8}}_{\sqrt{3}}\frac{1}{x\sqrt{1+x^{2}}}dx$= $=\frac{1}{2}\int^{\sqrt{8}}_{\sqrt{3}}\frac{(1+x^{2})'}{\sqrt{1+x^{2}}}dx$ $-\int^{\sqrt{8}}_{\sqrt{3}}\frac{(\frac{1}{x})'}{\sqrt{1+\frac{1}{x^{2}}}}dx =$ $\sqrt{1+x^{2}}|^{\sqrt{8}}_{\sqrt{3}}-ln(\frac{1}{x}+ \sqrt{1+\frac{1}{x^{2}}})$ $|^{\sqrt{8}}_{\sqrt{3}}$ $=1+\frac{1}{2}$$ln\frac{3}{2}$

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Use $x=\tan\theta$. Or, if you are familiar with hyperbolic functions, use $x=\sinh t$.

A somewhat less common substitution that works nicely is $x=\frac{1}{2}\left(t-\frac{1}{t}\right)$. Then $\sqrt{x^2+1}=\frac{1}{2}\left(t+\frac{1}{t}\right)$.

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You're doing great, now split this into three integrals; one by partial fractions.

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