integral Exercise asked in math group

Exercise:

Integrate, $\int{\frac{{{e}^{x}}\left( 2-{{x}^{2}} \right)}{\left( 1-x \right)\sqrt{1-{{x}^{2}}}}dx}$

Solution: we have ${{e}^{x}}\left( 2-{{x}^{2}} \right)={{e}^{x}}\left( 1+1-{{x}^{2}} \right)={{e}^{x}}+{{e}^{x}}\left( 1-{{x}^{2}} \right)$

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

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

But $\int{\frac{\left( 1+x \right){{e}^{x}}}{\sqrt{1-{{x}^{2}}}}dx}=\int{\frac{\left( 1+x \right){{e}^{x}}}{\sqrt{\left( 1-x \right)\left( 1+x \right)}}dx}=\int{\frac{\left( 1+x \right){{e}^{x}}}{\sqrt{1-x}.\sqrt{1+x}}dx}$

                         \[=\int{\frac{\left( 1+x \right){{e}^{x}}\sqrt{1+x}}{\left( 1+x \right)\sqrt{1-x}}dx}=\int{{{e}^{x}}\sqrt{\frac{1+x}{1-x}}}\,dx\]

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

Let $u={{e}^{x}}\,\,\And \,\,dv=\frac{dx}{\left( 1-x \right)\sqrt{1-{{x}^{2}}}}\Leftrightarrow du={{e}^{x}}\,\,\And \,\,\,v=\int{\frac{dx}{\left( 1-x \right)\sqrt{1-{{x}^{2}}}}}$

Put $x=\frac{t-1}{t}\Leftrightarrow dx=\frac{1}{{{t}^{2}}}dt$ and $1-x=1-\left( \frac{t-1}{t} \right)=\frac{t-t+1}{t}=\frac{1}{t}$

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

So $\int{\frac{dx}{\left( 1-x \right)\sqrt{1-{{x}^{2}}}}=\int{\frac{1}{\frac{1}{t}\sqrt{\frac{2t-1}{{{t}^{2}}}}}\times \frac{1}{{{t}^{2}}}dt}=\int{\frac{{{t}^{2}}}{\sqrt{2t-1}}\times \frac{dt}{{{t}^{2}}}=\int{\frac{dt}{\sqrt{2t-1}}}}}$

Let $w=2t-1\Leftrightarrow dw=2dt\Leftrightarrow dt=\frac{dw}{2}$  but $tx=t-1\Leftrightarrow t\left( x-1 \right)=-1$

So $\int{\frac{dt}{\sqrt{2t-1}}=\frac{1}{2}\int{\frac{dw}{\sqrt{w}}}={{w}^{1/2}}+c=\sqrt{2t-1}+c}$

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

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

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