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$$\int\frac{\sqrt {4x^2-9}}{x^2}dx$$

I tried to solve this using integration by parts, but I come up with something that is much more difficult to solve. How can this be solved?

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    $\begingroup$ The root is over $4$ only or the entire numerator??? $\endgroup$ – SchrodingersCat Nov 8 '15 at 11:37
  • $\begingroup$ The entire numerator $\endgroup$ – Erik Hambardzumyan Nov 8 '15 at 11:39
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$$\int \frac{\sqrt{4x^2-9}}{x^2}dx$$ $$=-\int \sqrt{4x^2-9} \,\ d(\frac{1}{x})$$ Using by parts, $$=-\left[\frac{\sqrt{4x^2-9}}{x}-\int \frac{4}{\sqrt{4x^2-9}}dx\right]$$ $$=-\left[\frac{\sqrt{4x^2-9}}{x}-2\int \frac{d(2x)}{\sqrt{(2x)^2-3^2}}\right]$$ $$=-\left[\frac{\sqrt{4x^2-9}}{x}-2 \ln \left|2x+\sqrt{4x^2-9}\right|\right]+c$$

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  • $\begingroup$ Nicely done Aniket. Once one can seethe function and its derivative, integration becomes so simple. +1 $\endgroup$ – Shailesh Nov 8 '15 at 13:53
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Trigonometric substitution There is another simpler method to solve the problem

Let $2x=3\sec\theta \implies dx=\frac{3}{2}\sec\theta \tan\theta \ d\theta$

$$\int \frac{\sqrt{4x^2-9}}{x^2}\ dx=\int \frac{\sqrt{9\sec^2\theta-9}}{\frac{9}{4}\sec^2\theta}\ \frac{3}{2}\sec\theta \tan\theta \ d\theta$$ taking positive value, $$=\frac{2}{3}\int \frac{3\tan\theta}{\sec\theta}\tan\theta \ d\theta$$ $$=2\int \frac{\tan^2 \theta}{\sec\theta}\ d\theta$$ $$=2\int \frac{\sec^2 \theta-1}{\sec\theta}\ d\theta$$ $$=2\int (\sec\theta-\cos\theta)\ d\theta$$

$$=2\left(\int \sec\theta\ d\theta-\int \cos\theta\ d\theta\right)$$

$$=2\left(\ln\left|\sec\theta+\tan\theta\right| -\sin\theta\right)+c$$ substituting $\sec\theta=\frac{2x}{3}$, one should get $$=2\ln|2x+\sqrt{4x^2-9}|-\frac{\sqrt{4x^2-9}}{x}+C$$

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  • $\begingroup$ $x^2 = \frac{9}{4} \sec^2 \theta$ I think you forgot the numerical part... Or am I missing something? $\endgroup$ – Gummy bears Nov 8 '15 at 12:29
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Hint. You may integrate by parts: $$ \int\frac{\sqrt {4x^2-9}}{x^2}dx=-\frac{\sqrt {4x^2-9}}{x}+8\int\frac1{\sqrt {4x^2-9}}dx $$ and you may conclude easily if you know that $$ (\text{arcsinh}\: (ax))'=\frac{a}{\sqrt {a^2x^2-1}}. $$

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  • $\begingroup$ C'est dommage qu'on ne puisse pas parler entre personnes sur ce site.. je lis régulièrement les réponses que tu as pu donner par le passer, je suis vraiment très impressionné! :) $\endgroup$ – ParaH2 Nov 8 '15 at 15:20
  • $\begingroup$ @Shadock Merci à toi pour tes mots sympas, cela va m'encourager :) $\endgroup$ – Olivier Oloa Nov 8 '15 at 16:55
  • $\begingroup$ Dommage qu'on ne puisse pas corriger ses fautes d'orthographe dans les commentaires à partir d'un certain temps! ^^ Quelle honte :P $\endgroup$ – ParaH2 Nov 8 '15 at 16:57
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By substitution: $\DeclareMathOperator\ach{arg\,cosh}$

Set $ x=\dfrac32\cosh t,\enspace t\ge0$, whence $\mathrm d\mkern1mu x=\dfrac32\sinh t\,\mathrm d\mkern1mu t$. With some hyperbolic trigonometry, we get \begin{align*} \int \frac{\sqrt{4x^2-9}}{x^2}\,\mathrm d\mkern1mu x&=2\int \frac{\sinh t}{\cosh^2t}\sinh t\,\mathrm d\mkern1mu t =2\int \tanh^2t\,\mathrm d\mkern1mu t \\&=2\int\bigl(1-(1-\tanh^2t)\bigr)\,\mathrm d\mkern1mu t=2(t-\tanh t)\\ &=2\Biggl(\ach\frac{2x}3-\frac{\sinh\bigl(\ach\frac{2x}3\bigr)}{\frac{2x}3}\Biggr)\\ &=2\Biggl(\ach\frac{2x}3-\frac{3\sqrt{\frac{4x^2}9-1}}{2x}\Biggr)=2\Biggl(\ach\frac{2x}3-\frac{\sqrt{4x^2-9}}{2x}\Biggr)\\ &=2\ln\bigl(2x+\sqrt{4x^2-9}\bigr)-\frac{\sqrt{4x^2-9}}{x}+\text{constant}. \end{align*}

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