A) \[{{e}^{x}}\cot x+c\]
B) \[-{{e}^{x}}\cot x+c\]
C) \[-{{e}^{x}}\tan x+c\]
D) \[{{e}^{x}}\tan x+c\]
Correct Answer: D
Solution :
\[\int_{{}}^{{}}{\left( \frac{2+\sin 2x}{1+\cos 2x} \right)\text{ }{{e}^{x}}\,dx}=\int_{{}}^{{}}{\left( \frac{2{{e}^{x}}}{1+\cos 2x} \right)dx}+\int_{{}}^{{}}{\frac{{{e}^{x}}\sin 2x}{1+\cos 2x}dx}\] \[=\int_{{}}^{{}}{{{e}^{x}}{{\sec }^{2}}x\,dx}+\int_{{}}^{{}}{{{e}^{x}}\tan x\,dx={{e}^{x}}\tan x+c}\].
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