Integrating $\int \sqrt{x+\sqrt{x^2+1}}\,\mathrm{d}x$ 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?

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

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

