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I'm struggling to find a suitable substitution for this integral:

$$ \int{\frac{\sqrt{x^2+4}}{x}} $$

I've tried $u=x^2+4$, $u^2=x^2+4$ and some trigonemetric identities, but not much progress. Can anybody help me figure out how to get to the answer:

$$ \sqrt{x^2+4}+\ln\left\vert\frac{\sqrt{x^2+4}-2}{\sqrt{x^2+4}+2}\right\vert + c $$

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    $\begingroup$ Did you try $x=2\tan u$? $\endgroup$
    – J.G.
    Nov 12, 2023 at 11:54
  • $\begingroup$ You could try $x=2\sinh u$ $\endgroup$ Nov 12, 2023 at 12:24
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    $\begingroup$ Hohner, actually, $u^2=x^2+4$ is a good substitution for your integral. Look at my answer. $\endgroup$
    – Angelo
    Nov 12, 2023 at 14:51

3 Answers 3

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Another way to calculate your integral.

By letting $\;t=\sqrt{x^2+4}\,,\,$ you get that

$x=\sqrt{t^2-4}\;$ and $\;\mathrm dx=\dfrac t{\sqrt{t^2-4}}\,\mathrm dt\,.$

Consequently ,

$\displaystyle\int\frac{\sqrt{x^2+4}}x\,\mathrm dx=\!\int\!\frac{t^2}{t^2-4}\,\mathrm dt=\!\int\!\left(1\!+\!\dfrac1{t-2}\!-\!\dfrac1{t+2}\right)\mathrm dt\!=$

$=t+\ln|t-2|-\ln|t+2|+c=$

$=t+\ln\left|\dfrac{t-2}{t+2}\right|+c=$

$=\sqrt{x^2+4}+\ln\left|\dfrac{\sqrt{x^2+4}-2}{\sqrt{x^2+4}+2}\right|+c\,.$

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Letting $x=2\tan u$, $\mathrm dx=2 \sec^2 u \,\mathrm du$, you get: $$\int \frac{\sqrt{x^2+4}}{x}\mathrm dx=2\int\frac{\sqrt{4\tan^2u+4}}{2\tan u} \sec^2 u\,\mathrm du=2\int \cot u \sec^3 u\,\mathrm du $$ Can you take it from here?

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  • $\begingroup$ Not entirely... According to Wolfram, $2\int \cot u \sec^3 u\,\mathrm du$ integrates to $2(\sec(u)+\log{\tan(\frac{u}{2})})$ Is that right? I don't think I've seen this form of integration, unless I'm missing something $\endgroup$
    – hohner
    Nov 12, 2023 at 14:52
  • $\begingroup$ @hohner yes, that's right. The answer may differ because of trig identities $\endgroup$ Nov 12, 2023 at 14:58
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$$\int \frac{\sqrt{4+ x^2}}{x}dx=\int \frac{{4+ x^2}}{x\sqrt{4+ x^2}}dx= \int \frac{{4}}{x\sqrt{4+ x^2}}dx+ \int \frac{{x}}{\sqrt{4+ x^2}}dx$$

$\int \frac{{x}}{\sqrt{4+ x^2}}dx$ is an easy integral $= \sqrt{4+ x^2} +C $ for $\int\frac{{4}}{x\sqrt{4+ x^2}}dx =\int\frac{{2 \times \frac{1}{2}}}{\frac{x}{2}\sqrt{1+ (\frac{x}{2})^2}}dx$ let $0.5 x = \tan t $ then $$ \int\frac{{2 \times \frac{1}{2}}}{\frac{x}{2}\sqrt{1+ (\frac{x}{2})^2}}dx= \int\frac{{2 \sec^2(t)}}{\tan(t)|\sec(t)|}dx = 2 \operatorname{sgn}(\cos(t))\int \csc(t) dt $$ $$=-2\ln \left|\csc(t) + \cot(t) \right| =-2\ln \left|\frac{\sqrt{x^2+4}}{x} + \frac{2}{x}\right| $$

so $$\int \frac{\sqrt{4+ x^2}}{x}dx =-2\ln \left|\frac{\sqrt{x^2+4}}{x} + \frac{2}{x}\right| +\sqrt{4+ x^2} +C $$

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