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