$\int_2^{\mathrm{e}}\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] d x=a+\frac{b}{\log 2}$, then

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Aicharya · Accepted Answer

$\mathrm{a}+\frac{\mathrm{b}}{\log 2}=\int_2^{\mathrm{e}}\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}ight] \mathrm{dx}$ Put $\log \mathrm{x}=\mathrm{t} \Rightarrow \frac{1}{\mathrm{x}} \mathrm{dx}=\mathrm{dt} \Rightarrow \mathrm{dx}=\mathrm{e}^{\mathrm{t}} \mathrm{dt}$ When $\mathrm{x}=\mathrm{e}, \mathrm{t}=1$ and when $\mathrm{x}=2, \mathrm{t}=\log 2$ $\begin{aligned} & 	herefore a+\frac{b}{\log 2}=\int_{\log 2}^1\left(\frac{1}{t}-\frac{1}{t^2}ight) e^t \cdot d t=\left[e^t \cdot \frac{1}{t}ight]_{\log 2}^1=e-\frac{e^{\log 2}}{\log 2}=e-\frac{2}{\log 2} \ & 	herefore a=e, b=-2 \end{aligned}$

$\int_2^{\mathrm{e}}\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] d x=a+\frac{b}{\log 2}$, then

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