A) \[x{{e}^{{{\tan }^{-1}}x}}+c\]
B) \[{{x}^{2}}{{e}^{{{\tan }^{-1}}x}}+c\]
C) \[\frac{1}{x}{{e}^{{{\tan }^{-1}}x}}+c\]
D) None of these
Correct Answer: A
Solution :
Putting \[{{\tan }^{-1}}x=t\] and \[\frac{dx}{1+{{x}^{2}}}=dt,\] we get \[\int_{{}}^{{}}{{{e}^{{{\tan }^{-1}}x}}\left( \frac{1+x+{{x}^{2}}}{1+{{x}^{2}}} \right)}\,dx=\int_{{}}^{{}}{{{e}^{t}}(\tan t+{{\sec }^{2}}t)\,dt}\] \[={{e}^{t}}\tan t+c=x\,{{e}^{{{\tan }^{-1}}x}}+c\] \[\left[ \text{Using }\int_{{}}^{{}}{{{e}^{x}}\left\{ f(x)+{f}'(x) \right\}dx={{e}^{x}}f(x)+C} \right]\].
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