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Integrating $$\int \sqrt{x+\sqrt{x^2+1}}\,\mathrm{d}x$$ Using substitution of $x=\tan \theta$, I got the required answer. But is there a more elegant solution to the problem?

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    $\begingroup$ Which is the measure of elegance? I mean, $x=\tan \theta$ works and is not particularly inelegant... $\endgroup$ – Miguel Aug 21 '15 at 20:21
  • $\begingroup$ You can solve it without substitution - just use complex numbers... $\endgroup$ – johannesvalks Aug 22 '15 at 0:53
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setting $$t=\sqrt{x+\sqrt{x^2+1}}$$ then we get after squaring $$t^2-x=\sqrt{x^2+1}$$ squaring again and solving for $x$ we get $$x=\frac{t^4-1}{2t^2}$$ and we obtain $$dx=\frac{t^4+1}{t^3}dt$$ and our integral will be $$\int t^2+\frac{1}{t^2}dt$$

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    $\begingroup$ Very nice substitution! $\endgroup$ – CivilSigma Aug 22 '15 at 0:51
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HINT....You could try $x=\sinh\theta$, because the square root expression simplifies dramatically

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  • $\begingroup$ excellent hint! $\endgroup$ – Math-fun Aug 21 '15 at 20:37
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HINT

$$ \sqrt{ x + \sqrt{ x^2 + 1 } } = \sqrt{ \frac{ x + i }{ 2 } } + \sqrt{ \frac{ x - i }{ 2 } } $$

That would be enough simple to solve the integral...

We get

$$ \begin{eqnarray} \int \sqrt{ x + \sqrt{ x^2 + 1 } } dx &=& \int \left\{ \sqrt{ \frac{ x + i }{ 2 } } + \sqrt{ \frac{ x - i }{ 2 } } \right\} d x\\\\ &=& \frac{4}{3} \left\{ \sqrt{ \frac{ x + i }{ 2 } }^3 + \sqrt{ \frac{ x - i }{ 2 } }^3 \right\} \quad \textrm{(*)}\\\\ &=& \bbox[16px,border:2px solid #800000] {\frac{4}{3} \sqrt{ x + \sqrt{ x^2 + 1 } } \left\{ x - \frac{1}{2} \sqrt{x^2+1}\right\}} \end{eqnarray} $$

(*) Where we have used

$$ a^3 + b^3 = \Big( a + b \Big) \Big( a^2 + b^2 - a b \Big) $$

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    $\begingroup$ @Dr.MV: The equation in the hint looks good to me. $(a+b)^2=a^2+b^2+\color{#C00000}{2}ab$ $\endgroup$ – robjohn Aug 21 '15 at 20:42
  • $\begingroup$ @Dr.MV: Nope, they are equal, as $( \sqrt{a+b} + \sqrt{a-b} )^2 = 2 a + 2 \sqrt{a^2-b^2}$ $\endgroup$ – johannesvalks Aug 21 '15 at 20:43
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    $\begingroup$ @johannesvalks Yes, it is. So, a big +1! $\endgroup$ – Mark Viola Aug 21 '15 at 20:45
  • $\begingroup$ @Dr.MV: Thanks ;) $\endgroup$ – johannesvalks Aug 21 '15 at 20:46
  • $\begingroup$ @robjohn Yes ... fancy math ... LOL $\endgroup$ – Mark Viola Aug 21 '15 at 20:46
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Let $$ I = \int \sqrt{x+\sqrt{x^2+1}} \, dx\;,$$ Let $$x+\sqrt{x^2+1} = t^2 \tag 1$$

Then $$ \left(1+\frac{x}{\sqrt{x^2+1}}\right) \, dx = 2t \, dt\Rightarrow t^2 \, dx = 2t\sqrt{x^2+1} \, dt$$

Now using $$\bullet\; \left(\sqrt{x^2+1}+x\right)\cdot \left(\sqrt{x^2+1}-x\right) = 1$$

So we get $$ \sqrt{x^2+1}-x = \frac{1}{\sqrt{x^2+1}+x} = \frac{1}{t^2} \tag 2$$

Now add $(1)$ and $(2)\;,$ we get $\displaystyle \sqrt{x^2+1} = \frac{t^2+t^{-2}}{2}$

So we get $$ dx = \frac{\left(t^2+t^{-2}\right)\cdot 2t}{2t^2} \, dt$$

so Integral $$\displaystyle I = \int ( t^2+t^{-2}) \, dt = \frac{t^3}{3}-\frac{1}{t}+\mathcal{C}$$

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