A) \[{{e}^{x}}\sqrt{\frac{1+x}{1-x}}+c\]
B) \[{{e}^{x}}\sqrt{1+x}+c\]
C) \[{{e}^{x}}\sqrt{1-x}+c\]
D) \[{{e}^{x}}\sqrt{\frac{1-x}{1+x}}+c\]
E) \[\sqrt{\frac{1+x}{1-x}}+c\]
Correct Answer: A
Solution :
Let \[I=\int{\frac{{{e}^{x}}(2-{{x}^{2}})dx}{(1-x)\sqrt{1-{{x}^{2}}}}}\] \[=\int{\frac{{{e}^{x}}(1-{{x}^{2}})}{(1-x)\sqrt{1-{{x}^{2}}}}dx}+\int{\frac{{{e}^{x}}}{(1-x)\sqrt{1-{{x}^{2}}}}}dx\] \[=\int{{{e}^{x}}}\sqrt{\frac{1+x}{1-x}}dx+\int{\frac{{{e}^{x}}}{(1-x)\sqrt{1-{{x}^{2}}}}}dx\] \[=\left[ {{e}^{x}}\sqrt{\frac{1+x}{1-x}}-\frac{1}{2}\int{\frac{d}{dx}}\sqrt{\frac{1+x}{1-x}}{{e}^{x}}dx \right]\] \[+\int{\frac{{{e}^{x}}}{(1-x)\sqrt{1-{{x}^{2}}}}dx}\] \[={{e}^{x}}\sqrt{\frac{1+x}{1-x}}-\int{\frac{{{e}^{x}}}{(1-x)\sqrt{1-{{x}^{2}}}}}dx\] \[+\int{\frac{{{e}^{x}}}{(1-x)\sqrt{1-{{x}^{2}}}}}dx\] \[={{e}^{x}}\sqrt{\frac{1+x}{1-x}}+c\]
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