A) \[4{{x}^{2}}+12x+6\log x-\frac{1}{x}+c\]
B) \[4{{x}^{2}}+12x-6\log x-\frac{2}{x}+c\]
C) \[2{{x}^{2}}+8x+3\log x-\frac{2}{x}+c\]
D) \[8{{x}^{2}}+6x+6\log x+\frac{2}{x}+c\]
Correct Answer: A
Solution :
\[\int_{{}}^{{}}{\frac{1}{{{x}^{2}}}{{(2x+1)}^{3}}dx=\int_{{}}^{{}}{\frac{(8{{x}^{3}}+1+12{{x}^{2}}+6x)}{{{x}^{2}}}\,dx}}\] \[=\int_{{}}^{{}}{\left( 8x+12+\frac{6}{x}+\frac{1}{{{x}^{2}}} \right)\,dx}=4{{x}^{2}}+12x+6\log x-\frac{1}{x}+c\].
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