Integral of Nested Radicals

Exercise:

Integrate, $\int{\frac{dx}{\sqrt{2+\sqrt{1+\sqrt{x}}}}}$

Solution: Let ${{w}^{2}}=2+\sqrt{1+\sqrt{x}}\Rightarrow {{\left( {{w}^{2}}-2 \right)}^{2}}=1+\sqrt{x}$$\Rightarrow x={{\left[ {{\left( {{w}^{2}}-2 \right)}^{2}}-1 \right]}^{2}}$

So $dx=2\left[ {{\left( {{w}^{2}}-2 \right)}^{2}}-1 \right]\left( 2\left( {{w}^{2}}-2 \right)2w \right)dw$ $=8w\left[ {{\left( {{w}^{2}}-2 \right)}^{2}}-1 \right]\left( {{w}^{2}}-2 \right)dw$

So $\int{\frac{dx}{\sqrt{2+\sqrt{1+\sqrt{x}}}}=\int{\frac{8w\left( {{\left( {{w}^{2}}-2 \right)}^{2}}-1 \right)\left( {{w}^{2}}-2 \right)dw}{\sqrt{{{w}^{2}}}}}}$

$=8\int{\left( {{\left( {{w}^{2}}-2 \right)}^{2}}-1 \right)\left( {{w}^{2}}-2 \right)dw}$

But $\left[ {{\left( {{w}^{2}}-2 \right)}^{2}}-1 \right]\left( {{w}^{2}}-2 \right)=\left[ {{w}^{4}}-4{{w}^{2}}+4-1 \right]\left( {{w}^{2}}-2 \right)$

 $={{w}^{6}}-6{{w}^{4}}+11{{w}^{2}}-6$

So $\int{\frac{dx}{\sqrt{2+\sqrt{1+\sqrt{x}}}}=8\int{{{w}^{6}}dw}-48\int{{{w}^{4}}dw+88\int{{{w}^{2}}dw-48\int{dw}}}}$

                                 $=\frac{8}{7}{{w}^{7}}-\frac{48}{5}{{w}^{5}}+\frac{88}{3}{{w}^{3}}-48w+c$

$=\frac{8}{7}{{\left( \sqrt{2+\sqrt{1+\sqrt{x}}} \right)}^{7}}-\frac{48}{5}{{\left( \sqrt{2+\sqrt{1+\sqrt{x}}} \right)}^{5}}+\frac{88}{3}{{\left( \sqrt{2+\sqrt{1+\sqrt{x}}} \right)}^{3}}-48\sqrt{2+\sqrt{1+\sqrt{x}}}+c$