A) \[2\log 2-\frac{1}{6}\]
B) \[\log \frac{16}{9}-\frac{1}{6}\]
C) \[\log \frac{4}{3}-\frac{1}{6}\]
D) \[\log \frac{16}{9}+\frac{1}{6}\]
Correct Answer: B
Solution :
\[I=\int_{2}^{3}{\frac{x+1}{{{x}^{2}}(x-1)}}dx=\int_{2}^{3}{\left( \frac{A}{{{x}^{2}}}+\frac{B}{x}+\frac{C}{x-1} \right)}\,dx\] \[A(x-1)+B(x)(x-1)+C({{x}^{2}})=x+1\] Put \[x=0,\,1,\,-1,\]we get \[A=-1,B=-2,C=2\] Þ \[I=-\int_{2}^{3}{\frac{dx}{{{x}^{2}}}-2\int_{2}^{3}{\frac{dx}{x}+2\int_{2}^{3}{\frac{dx}{x-1}}}}\] Þ \[I=\left[ \frac{1}{x} \right]_{2}^{3}-2[\log x]_{2}^{3}+2[\log (x-1)]_{2}^{3}\] Þ \[I=\frac{1}{3}-\frac{1}{2}-2\log \frac{3}{2}+2\log 2\]Þ \[I=\log \frac{16}{9}-\frac{1}{6}\].
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