Integral exercise asked by teacher حاجم الربيعي in the mathematics teacher group ( Great idea )

Exercise:

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

Solution: Let $f\left( x \right)=\frac{p\left( x \right)}{q\left( x \right)}$ where $p\left( x \right)=\sqrt{x\sqrt{x\sqrt{x....}}}\,\,\,\And \,\,\,q\left( x \right)=\sqrt{x+\sqrt{x+\sqrt{x+....}}}$

So $p\left( x \right)=\sqrt{x\sqrt{x\sqrt{x....}}}={{\left( x{{\left( x{{\left( {{x}^{\frac{1}{2}}} \right)}^{}} \right)}^{1/2}} \right)}^{1/2}}={{x}^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....}}={{x}^{\sum\limits_{i=1}^{\infty }{{{\left( \frac{1}{2} \right)}^{i}}}}}={{x}^{\frac{\frac{1}{2}}{1-\frac{1}{2}}}}={{x}^{1}}=x$

Let $y=\sqrt{x+\sqrt{x+\sqrt{x+....}}}\Leftrightarrow y=\sqrt{x+y}\Leftrightarrow {{y}^{2}}=x+y$

So $\frac{d}{dx}\left( {{y}^{2}} \right)=\frac{d}{dx}\left( x+y \right)\Leftrightarrow 2y\frac{dy}{dx}=1+\frac{dy}{dx}\Leftrightarrow 2ydy=dx+dy\Leftrightarrow \left( 2y-1 \right)dy=dx$

Thus $\int{\frac{\sqrt{x\sqrt{x\sqrt{x......}}}}{\sqrt{x+\sqrt{x+\sqrt{x+...}}}}dx}=\int{\frac{{{y}^{2}}-y}{y}\left( 2y-1 \right)dy}=\int{\frac{y\left( y-1 \right)\left( 2y-1 \right)}{y}}\,dy$

$=\int{\left( y-1 \right)\left( 2y-1 \right)dy}=\int{\left( 2{{y}^{2}}-3y+1 \right)dy=\int{2{{y}^{2}}dy-3\int{y}dy+\int{dy}}}$

$=\frac{2}{3}{{y}^{3}}-\frac{3}{2}{{y}^{2}}+y+c=\frac{2}{3}{{\left( \sqrt{x+\sqrt{x+...}} \right)}^{3}}-\frac{3}{2}{{\left( \sqrt{x+\sqrt{x+....}} \right)}^{2}}+\sqrt{x+\sqrt{x+....}}+c$