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

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 $$\int t^2+\frac{1}{t^2}dt$$

HINT....You could try $x=\sinh\theta$, because the square root expression simplifies dramatically

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

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