Integral Calculus Question 369
Question: $ \int_{{}}^{{}}{\frac{{e^{{{\tan }^{-1}}x}}}{1+x^{2}}dx=} $
[MP PET 1987]
Options:
A) $ \log (1+x^{2})+c $
B) $ \log {e^{{{\tan }^{-1}}x}}+c $
C) $ {e^{{{\tan }^{-1}}x}}+c $
D) $ {{\tan }^{-1}}{e^{{{\tan }^{-1}}x}}+c $
Show Answer
Answer:
Correct Answer: C
Solution:
Putting $ t={{\tan }^{-1}}x\Rightarrow dt=\frac{1}{1+x^{2}},dx, $ we get $ \int_{{}}^{{}}{\frac{{e^{{{\tan }^{-1}}x}}}{1+x^{2}},dx}=\int_{{}}^{{}}{e^{t}dt} $ $ =e^{t}+c={e^{{{\tan }^{-1}}x}}+c. $
BETA
