Integral Calculus Question 157
Question: $ \int_{{}}^{{}}{{e^{{{\tan }^{-1}}x}}}( \frac{1+x+x^{2}}{1+x^{2}} )\ dx $ is equal to
Options:
A) $ x{e^{{{\tan }^{-1}}x}}+c $
B) $ x^{2}{e^{{{\tan }^{-1}}x}}+c $
C) $ \frac{1}{x}{e^{{{\tan }^{-1}}x}}+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Putting $ {{\tan }^{-1}}x=t $ and $ \frac{dx}{1+x^{2}}=dt, $ we get $ \int_{{}}^{{}}{{e^{{{\tan }^{-1}}x}}( \frac{1+x+x^{2}}{1+x^{2}} )},dx=\int_{{}}^{{}}{e^{t}(\tan t+{{\sec }^{2}}t),dt} $ $ =e^{t}\tan t+c=x,{e^{{{\tan }^{-1}}x}}+c $ $ [ \text{Using }\int_{{}}^{{}}{e^{x}{ f(x)+{f}’(x) }dx=e^{x}f(x)+C} ] $ .
BETA
