A) \[\log ({{x}^{2}}+\sin 2x+2x)+c\]
B) \[-\log ({{x}^{2}}+\sin 2x+2x)+c\]
C) \[\frac{1}{2}\log ({{x}^{2}}+\sin 2x+2x)+c\]
D) None of these
Correct Answer: C
Solution :
Put \[{{x}^{2}}+\sin 2x+2x=t,\] then it reduces to \[\frac{1}{2}\int_{{}}^{{}}{\frac{1}{t}\,dt}=\frac{1}{2}\log t+c=\frac{1}{2}\log ({{x}^{2}}+\sin 2x+2x)+c.\]You need to login to perform this action.
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