Exercise:
Let $f:\left( -1,1 \right)\to \mathbb{R}$ be a function defined to be $f\left( x \right)=\frac{1+x}{\sqrt{1-{{x}^{2}}}}$
1) Show that, $\frac{d}{dx}\left( f\left( x \right) \right)=\frac{1}{\left( 1-x \right)\sqrt{1-{{x}^{2}}}}$
2) Show that, $\int{\frac{{{e}^{x}}\left( 2-{{x}^{2}} \right)}{\left( 1-x \right)\sqrt{1-{{x}^{2}}}}dx}={{e}^{x}}f\left( x \right)+c$
Solution:
1) Observe that, $f\left( x \right)=\frac{1+x}{\sqrt{1-{{x}^{2}}}}\Leftrightarrow f'\left( x \right)=\frac{x\left( x+1 \right)}{\left( 1-{{x}^{2}} \right)\sqrt{1-{{x}^{2}}}}+\frac{1}{\sqrt{1-{{x}^{2}}}}=\frac{1}{\left( 1-x \right)\sqrt{1-{{x}^{2}}}}$
2) we have ${{e}^{x}}\left( 2-{{x}^{2}} \right)={{e}^{x}}\left( 1+1-{{x}^{2}} \right)={{e}^{x}}+{{e}^{x}}\left( 1-{{x}^{2}} \right)={{e}^{x}}+{{e}^{x}}\left( 1-x \right)\left( 1+x \right)$
So $\int{\frac{{{e}^{x}}\left( 2-{{x}^{2}} \right)}{\left( 1-x \right)\sqrt{1-{{x}^{2}}}}dx}=\int{\frac{{{e}^{x}}+{{e}^{x}}\left( 1-x \right)\left( 1+x \right)}{\left( 1-x \right)\sqrt{1-{{x}^{2}}}}dx}$
$=\int{\frac{{{e}^{x}}}{\left( 1-x \right)\sqrt{1-{{x}^{2}}}}dx+\int{\frac{{{e}^{x}}\left( 1+x \right)}{\sqrt{1-{{x}^{2}}}}dx}}$
So $\int{\frac{{{e}^{x}}\left( 2-{{x}^{2}} \right)}{\left( 1-x \right)\sqrt{1-{{x}^{2}}}}dx}=\int{{{e}^{x}}\left( f'\left( x \right)+f\left( x \right) \right)dx}=\int{{{e}^{x}}f'\left( x \right)dx+\int{{{e}^{x}}f\left( x \right)dx}}$
Let $u={{e}^{x}}\,\,\And \,\,dv=df\left( x \right)\Leftrightarrow du={{e}^{x}}dx\,\,\And \,\,v=f\left( x \right)$
So $\int{{{e}^{x}}f'\left( x \right)dx}=f\left( x \right){{e}^{x}}-\int{{{e}^{x}}f\left( x \right)dx}$
Thus $\int{{{e}^{x}}f'\left( x \right)dx}+\int{{{e}^{x}}f\left( x \right)dx}=f\left( x \right){{e}^{x}}-\int{{{e}^{x}}f\left( x \right)dx+\int{{{e}^{x}}f\left( x \right)dx}=f\left( x \right){{e}^{x}}+c}$
Therefore , $\int{\frac{{{e}^{x}}\left( 2-{{x}^{2}} \right)}{\left( 1-x \right)\sqrt{1-{{x}^{2}}}}dx}={{e}^{x}}\left( \frac{1+x}{\sqrt{1-{{x}^{2}}}} \right)+c$
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