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