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