A) \[{{\log }_{e}}x+c\]
B) \[{{\log }_{e}}x+2{{\tan }^{-1}}x+c\]
C) \[{{\log }_{e}}\frac{1}{{{x}^{2}}+1}+c\]
D) \[{{\log }_{e}}\{x({{x}^{2}}+1)\}+c\]
Correct Answer: B
Solution :
\[\int{\frac{{{(x+1)}^{2}}}{x({{x}^{2}}+1)}dx}\]\[=\int{\frac{{{x}^{2}}+1+2x}{x({{x}^{2}}+1)}dx}\] \[=\int{\frac{{{x}^{2}}+1}{x({{x}^{2}}+1)}dx+2\int{\frac{x}{x({{x}^{2}}+1)}dx}}\] \[=\int{\frac{dx}{x}+2\int{\frac{dx}{{{x}^{2}}+1}}}={{\log }_{e}}x+2{{\tan }^{-1}}x+c\].
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