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