# to compute $\int_{2}^{3}\sqrt{1+\frac{1}{x^{2}}} \: \: dx$

Help me please to compute: $\int_{2}^{3}\sqrt{1+\frac{1}{x^{2}}} \: \: dx$

Thanks a lot!

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Hint: let $x=\tan{\theta}$, $dx=\sec^2{\theta} d \theta$:

\begin{align} \int_{2}^{3} dx \: \sqrt{1+\frac{1}{x^{2}}} &= \int_{\arctan{2}}^{\arctan{3}} \frac{d \theta}{\sin{\theta} \cos^2{\theta}} \\ &= \int_{\arctan{2}}^{\arctan{3}} \frac{d \theta \, \sin{\theta}}{\sin^2{\theta} \cos^2{\theta}} \end{align}

Now let $y=\cos{\theta}$, $dy=-\sin{\theta} d \theta$:

\begin{align} \int_{2}^{3} dx \: \sqrt{1+\frac{1}{x^{2}}} &= \int_{\frac{1}{\sqrt{10}}}^{\frac{1}{\sqrt{5}}} \frac{dy}{y^2 (1-y^2)} \\ &= \int_{\frac{1}{\sqrt{10}}}^{\frac{1}{\sqrt{5}}} dy \: \left ( \frac{1}{y^2} + \frac{1}{1-y^2} \right )\\ &= \int_{\frac{1}{\sqrt{10}}}^{\frac{1}{\sqrt{5}}} dy \: \left [ \frac{1}{y^2} + \frac{1}{2} \left ( \frac{1}{1-y} + \frac{1}{1+y} \right ) \right ] \\ \end{align}

I think you can take it from here.

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We let $x=1/y$ that yields $$-\int_{1/2}^{1/3}\frac{\sqrt{1+y^2}}{y^2} dy=$$ $$\int_{1/2}^{1/3}\left(\frac{1}{y}\right)'\sqrt{1+y^2} dy=$$ $$\left[\frac{\sqrt{1+y^2}}{y}\right]_{1/2}^{1/3}-\int_{1/2}^{1/3} \frac{1}{\sqrt{1+y^2}} dy=\sqrt{10}-\sqrt{5} -\sinh^{-1}(1/3)+\sinh^{-1}(1/2)$$

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Hint : set $x = f(y)$ where is $f$ is a very well-known function.

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Not informative and questionable hint. For example $x=sin(y)$ and it gives no way out in finding the solution. Could you please elaborate more on how your hint could/should help? –  Tomas Jan 28 '13 at 12:50