A) \[a=e,\,\,b=-2\]
B) \[a=e,\,\,b=2\]
C) \[a=-e,\,\,b=2\]
D) None of these
Correct Answer: A
Solution :
Given, \[\int_{2}^{e}{\left( \frac{1}{\log x}-\frac{1}{{{(\log x)}^{2}}} \right)dx=a+\frac{b}{\log 2}}\] put\[\log x=z\] \[\Rightarrow \] \[x={{e}^{z}}\] \[\Rightarrow \] \[dx={{e}^{z}}dz\] \[\therefore \] \[\int_{2}^{e}{\left( \frac{1}{\log x}-\frac{1}{{{(\log x)}^{2}}} \right)dx}\] \[=\int_{\log 2}^{1}{\left( \frac{1}{z}-\frac{1}{{{z}^{2}}} \right){{e}^{z}}dz}\] \[=\int_{\log 2}^{1}{{{e}^{z}}\left( \frac{1}{z}+d\left( \frac{1}{z} \right) \right)dz}\] \[=\left[ {{e}^{z}}\cdot \frac{1}{z} \right]_{\log 2}^{1}=e-\frac{2}{\log 2}\] \[\therefore \] \[a=e\]and\[b=-2\]
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