Integral exercise asked in the mathematics teacher group Level 3

Exercise:

Integrate, $\int{\frac{x+1}{\left( x-1 \right)\sqrt{{{x}^{3}}+x}}dx}$

Solution: we have $\sqrt{{{x}^{3}}+x}=\sqrt{{{x}^{2}}\left( x+\frac{x}{{{x}^{2}}} \right)}=x\sqrt{x+\frac{1}{x}}$

So $\frac{x+1}{\left( x-1 \right)\sqrt{{{x}^{3}}+x}}=\frac{x+1}{\left( x-1 \right)x\sqrt{x+\frac{1}{x}}}\times \frac{x-1}{x-1}=\frac{{{x}^{2}}-1}{{{\left( x-1 \right)}^{2}}x\sqrt{x+\frac{1}{x}}}$

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

So $\int{\frac{x+1}{\left( x-1 \right)\sqrt{{{x}^{3}}+x}}dx}=\int{\frac{1-\frac{1}{{{x}^{2}}}}{\left( x+\frac{1}{x}-2 \right)\sqrt{x+\frac{1}{x}}}dx}$

Let $u=x+\frac{1}{x}\Leftrightarrow du=\left( 1-\frac{1}{{{x}^{2}}} \right)dx$

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

Let ${{w}^{2}}=u\Leftrightarrow 2wdw=du$

So $\int{\frac{du}{\left( u-2 \right)\sqrt{u}}=\int{\frac{2wdw}{\left( {{w}^{2}}-2 \right)w}}=\int{\frac{2dw}{{{w}^{2}}-2}=2\int{\frac{dw}{{{w}^{2}}-{{\left( \sqrt{2} \right)}^{2}}}}}}$

Let $w=\sqrt{2}\,t\Leftrightarrow dw=\sqrt{2}\,dt$

So $\int{\frac{dw}{{{w}^{2}}-2}=\int{\frac{\sqrt{2}dt}{2{{t}^{2}}-2}=\frac{\sqrt{2}}{2}\int{\frac{dt}{{{t}^{2}}-1}=\frac{\sqrt{2}}{2}\operatorname{arctanh}\left( t \right)+c}}}=\frac{\sqrt{2}}{2}\operatorname{arctanh}\left( \frac{\sqrt{u}}{\sqrt{2}} \right)+c$

Thus$\int{\frac{x+1}{\left( x-1 \right)\sqrt{{{x}^{3}}+x}}dx=\frac{\sqrt{2}}{2}}\operatorname{arctanh}\left( \sqrt{\frac{{{x}^{2}}+1}{2}} \right)+c$