A) \[f(x)=\frac{1}{2}{{x}^{2}}\]
B) \[g(x)=\log x\]
C) \[L=1\]
D) None of these
Correct Answer: D
Solution :
[d] \[\int{x\,\,\log \left( 1+\frac{1}{x} \right)dx}\] \[=\log \left( 1+\frac{1}{x} \right).\frac{{{x}^{2}}}{2}-\int{\frac{x}{x+1}\cdot \left( -\frac{1}{{{x}^{2}}} \right)\cdot \frac{{{x}^{2}}}{2}dx}\] \[=\frac{{{x}^{2}}}{2}\log \left( \frac{x+1}{x} \right)\cdot \frac{{{x}^{2}}}{2}+\frac{1}{2}\int{\frac{x+1-1}{x+1}}dx\] \[=\frac{{{x}^{2}}}{2}\log \left( \frac{x+1}{x} \right)+\frac{1}{2}x-\frac{1}{2}\log (x+1)+c\] \[=\left( \frac{{{x}^{2}}-1}{2} \right)\log (x+1)-\frac{{{x}^{2}}}{2}\log x+\frac{1}{2}x+c\]
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