A) \[x\log 2x-\frac{{{x}^{2}}}{2}\]
B) \[x\log 2x-\frac{x}{2}\]
C) \[{{x}^{2}}\log 2x-\frac{x}{2}\]
D) \[x\log 2x-x+c\]
Correct Answer: D
Solution :
\[I=\int_{{}}^{{}}{\log 2xdx}=\int_{{}}^{{}}{\log 2x.1dx}\] Using Integration by parts \[I=\log 2x.x-\int_{{}}^{{}}{\frac{2}{2x}.\int_{{}}^{{}}{1.dx}}\] \[=x\log 2x-\int_{{}}^{{}}{\frac{1}{x}xdx+c=x\log 2x-x+c}\]
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