A) \[\log \frac{16}{9}+\frac{1}{6}\]
B) \[\log \frac{16}{9}-\frac{1}{6}\]
C) \[2\log 2-\frac{1}{6}\]
D) \[\log \frac{4}{3}-\frac{1}{6}\]
Correct Answer: B
Solution :
Let \[I=\int_{2}^{3}{\frac{x+1}{{{x}^{2}}(x-1)}dx}\] \[=\int_{2}^{3}{\left( \frac{-2}{x}-\frac{1}{{{x}^{2}}}+\frac{2}{x-1} \right)dx}\] \[=\left[ -2\log x+\frac{1}{x}+2\log (x-1) \right]_{2}^{3}\] \[=\left[ 2\log \left( \frac{x-1}{x} \right)+\frac{1}{x} \right]_{2}^{3}\] \[=\left[ 2\left( \log \frac{2}{3}-\log \frac{1}{2} \right)+\frac{1}{3}-\frac{1}{2} \right]\] \[=2\log \frac{4}{3}-\frac{1}{6}\] \[=\log \frac{16}{9}-\frac{1}{6}\]You need to login to perform this action.
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