Integral Calculus Question 454
Question: The value of $ \int_{{}}^{{}}{( 1+\frac{1}{x^{2}} )\ {e^{( x-\frac{1}{x} )}}}\ dx $ equals
[Kurukshetra CEE 1998]
Options:
A) $ {e^{x-\frac{1}{x}}}+c $
B) $ {e^{x+\frac{1}{x}}}+c $
C) $ {e^{x^{2}-\frac{1}{x}}}+c $
D) $ {e^{x^{2}+\frac{1}{x^{2}}}}+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int_{{}}^{{}}{( 1+\frac{1}{x^{2}} ){e^{x-\frac{1}{x}}}dx} $ . Put $ x-\frac{1}{x}=t\Rightarrow ( 1+\frac{1}{x^{2}} ),dx=dt $
$ \therefore ,I=\int_{{}}^{{}}{e^{t}dt=e^{t}+c={e^{x-\frac{1}{x}}}+c} $ .
BETA
