মান निर्ণয় क(রো : $$ \int\left\{\frac{1}{\log x}-\frac{1}{(\log x)^{2}}\right\} d x $$

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1. Let 
   $$
   u=\log x.
   $$
   Then, 
   $$
   \frac{du}{dx}=\frac{1}{x} \quad \text{or} \quad dx=x\,du=e^u\,du.
   $$

2. Substitute into the integral:
   $$
   \int\left(\frac{1}{\log x}-\frac{1}{(\log x)^2}\right)dx
   =\int\left(\frac{1}{u}-\frac{1}{u^2}\right)e^u\,du.
   $$
   That is,
   $$
   \int\frac{e^u}{u}\,du-\int\frac{e^u}{u^2}\,du.
   $$

3. Notice that the derivative of 
   $$
   \frac{e^u}{u}
   $$
   is computed as follows:
   $$
   \frac{d}{du}\left(\frac{e^u}{u}\right)
   =\frac{e^u\cdot u - e^u\cdot 1}{u^2}
   =\frac{e^u(u-1)}{u^2}.
   $$
   Thus,
   $$
   \frac{d}{du}\left(\frac{e^u}{u}\right)
   =\frac{e^u}{u}-\frac{e^u}{u^2}.
   $$

4. Since the integrand matches the derivative, we have:
   $$
   \int\left(\frac{e^u}{u}-\frac{e^u}{u^2}\right)du
   =\frac{e^u}{u}+C.
   $$

5. Re-substitute $$ u=\log x $$ and $$ e^u=x $$:
   $$
   \frac{e^u}{u}=\frac{x}{\log x}.
   $$

6. Therefore, the evaluated integral is:
   $$
   \frac{x}{\log x}+C.
   $$
The integral evaluates to $$ \frac{x}{\log x} + C $$.

মান निर्ণয় क(রো : \( \int\left\{\frac{1}{\log x}-\frac{1}{(\log x)^{2}}\right\} d x \)

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