Integral-Calculus Question 514
Question: The value of $ \int{{e^{ta{n^{-1}}}}^{x}\frac{(1+x+x^{2})}{1+x^{2}}dx} $ is
Options:
A) $ x{e^{{{\tan }^{-1}}}}x+c $
B) $ {{\tan }^{-1}}x+C $
C) $ {e^{{{\tan }^{-1}}x}}+2x+C $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Put $ x=\tan \theta \Rightarrow dx={{\sec }^{2}}\theta d\theta $ $ I=\int{{e^{\theta }}\frac{1+\tan \theta +{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }.{{\sec }^{2}}\theta d\theta } $ $ =\int{{e^{\theta }}(tan\theta +sec^{2}\theta )d\theta } $ $ ={e^{\theta }}\tan \theta +c=x{e^{{{\tan }^{-1}}x}}+c $
BETA
