Integral Calculus Question 318
Question: $ \int_{{}}^{{}}{\frac{e^{x}(x+1)}{{{\cos }^{2}}(xe^{x})}dx=} $
[Roorkee 1979; MP PET 1995; Pb. CET 2001]
Options:
A) $ \tan (xe^{x})+c $
B) $ \sec (xe^{x})\tan (xe^{x})+c $
C) $ -\tan (xe^{x})+c $
D) None of these
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Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{e^{x}(x+1)}{{{\cos }^{2}}(xe^{x})}=\int_{{}}^{{}}{e^{x}(x+1){{\sec }^{2}}(xe^{x})dx}} $ Putting $ xe^{x}=t\Rightarrow (x+1)e^{x}dx=dt $ , we get $ \int_{{}}^{{}}{{{\sec }^{2}}t,dt=\tan t+c}=\tan (xe^{x})+c. $
BETA
