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?

Solution 1:

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

Solution 2:

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

Solution 3:

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

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

Solution 4:

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