A) \[{{e}^{m+{{\tan }^{-1}}x}}\]
B) \[{{e}^{{{\tan }^{-1}}x}}\]
C) \[\frac{1}{m}{{\tan }^{-1}}x\]
D) None of these
Correct Answer: D
Solution :
Let \[I=\int{\frac{{{e}^{m\,{{\tan }^{-1}}x}}}{1+{{x}^{2}}}}dx\] Let \[{{\tan }^{-1}}x=t\] \[\Rightarrow \] \[\frac{1}{1+{{x}^{2}}}dx=dt\] \[\therefore \] \[I=\int{{{e}^{mt}}dt}=\frac{{{e}^{mt}}}{m}+c\] \[=\frac{{{e}^{m{{\tan }^{-1}}x}}}{m}+c\]
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