A) \[\log (x+\log \sec x)+c\]
B) \[-\log (x+\log \sec x)+c\]
C) \[\log (x-\log \sec x)+c\]
D) None of these
Correct Answer: A
Solution :
Put \[t=x+\log \sec x\Rightarrow dt=(1+\tan x)\,dx,\] then \[\int_{{}}^{{}}{\frac{1+\tan x}{x+\log \sec x}\,dx=\int_{{}}^{{}}{\frac{1}{t}\,dt=\log t+c}}\] \[=\log (x+\log \sec x)+c.\]
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