A) \[\tan \,(1+\log x)+c\]
B) \[\cot \,(1+\log x)+c\]
C) \[-\tan \,(1+\log x)+c\]
D) \[-\cot (\,1+\log x)+c\]
Correct Answer: A
Solution :
Put \[1+\log x=t\Rightarrow \frac{1}{x}\,dx=dt,\] then \[\int_{{}}^{{}}{\frac{1}{x{{\cos }^{2}}(1+\log x)}\,dx}=\int_{{}}^{{}}{\frac{dt}{{{\cos }^{2}}t}=\int_{{}}^{{}}{{{\sec }^{2}}t\,dt}}\] \[=\tan t+c=\tan (1+\log x)+c.\]
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