Integral Exercise -Substitution Tech

Exercise:

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

Solution: $u=\sqrt{2+\sqrt{1+\sqrt{x}}}\Leftrightarrow {{u}^{2}}=2+\sqrt{1+\sqrt{x}}\Leftrightarrow {{\left( {{u}^{2}}-2 \right)}^{2}}=1+\sqrt{x}\Leftrightarrow {{\left( {{\left( {{u}^{2}}-2 \right)}^{2}}-1 \right)}^{2}}=x$

${{\left( {{u}^{2}}-2-1 \right)}^{2}}{{\left( {{u}^{2}}-2+1 \right)}^{2}}=x\Leftrightarrow {{\left( {{u}^{2}}-3 \right)}^{2}}{{\left( {{u}^{2}}-1 \right)}^{2}}=x$

 So $dx=8u\left( {{\left( {{u}^{2}}-2 \right)}^{2}}-1 \right)\left( {{u}^{2}}-2 \right)du$

Thus \(\int{\frac{dx}{\sqrt{2+\sqrt{1+\sqrt{x}}}}=\int{\frac{8u\left( {{\left( {{u}^{2}}-2 \right)}^{2}}-1 \right)\left( {{u}^{2}}-2 \right)}{u}du}}\) $=8\int{\left( {{\left( {{u}^{2}}-2 \right)}^{2}}-1 \right)\left( {{u}^{2}}-2 \right)du}$

But $\left( {{\left( {{u}^{2}}-2 \right)}^{2}}-1 \right)\left( {{u}^{2}}-2 \right)={{\left( {{u}^{2}}-2 \right)}^{3}}-\left( {{u}^{2}}-2 \right)={{u}^{6}}-6{{u}^{4}}+11{{u}^{2}}-6$

So $\int{\left( {{\left( {{u}^{2}}-2 \right)}^{2}}-1 \right)\left( {{u}^{2}}-2 \right)du}=\int{\left( {{u}^{6}}-6{{u}^{4}}+11{{u}^{2}}-6 \right)du=\frac{{{u}^{7}}}{7}-\frac{6}{5}{{u}^{5}}+\frac{11}{3}{{u}^{3}}-6u+c}$

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

$-48\sqrt{2+\sqrt{1+\sqrt{x}}}+c$