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I'm having trouble solving the following integral $$ L = \int_{1}^{e}\sqrt{1 + \frac{1}{x^2}}dx = \int_{1}^{e}\frac{1}{x}\sqrt{1 + x^2}dx $$

Concerning the length of $\ln x$, $x = 1$ to $x = e$. Making $x = \tan \theta$, we obtain $$ F(\theta) = \int \frac{\sec^3 \theta}{\tan \theta}d\theta = ? $$

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  • $\begingroup$ Answer: $F(x) = \sqrt{1 + x^2} + \frac{1}{2}\ln(\frac{\sqrt{1 + x^2} - 1}{\sqrt{1 + x^2} + 1}) + C$. $\endgroup$ – Mathsource Sep 25 '16 at 18:06
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    $\begingroup$ $x=sinh (u)$, maybe? $\endgroup$ – Oscar Lanzi Sep 25 '16 at 18:08
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Hint: From where you left off,

$${\sec^3\theta\over\tan\theta}={1\over\sin\theta\cos^2\theta}={\sin\theta\over(1-\cos^2\theta)\cos^2\theta}$$

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Hint

$$\int \limits^{e}_{1}\sqrt{1+\frac{1}{x^{2}} } dx=\int \limits^{e}_{1}\frac{\sqrt{1+x^{2}} }{x} dx$$ Then use the substitution $\sqrt{1+x^{2}} = y \implies dx=\frac{y}{\sqrt{y^{2}-1} } dy$ . To get the following integral $$\int \limits^{\sqrt{1+e^{2}} }_{\sqrt{2} }\frac{y^{2}}{y^{2}-1} dy$$ Which is easy to integrat.

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