Integral Calculus Question 473
Question: $ \int_{{}}^{{}}{\frac{{e^{m{{\tan }^{-1}}x}}}{1+x^{2}}dx} $ equals to
[RPET 2001]
Options:
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
Show Answer
Answer:
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 $ .
BETA
