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If $\mathrm{f}(\mathrm{x})=\frac{1}{\log \mathrm{x}}, \mathrm{g}(\mathrm{x})=\frac{1}{(\log \mathrm{x})^2}$
(Where C is a constant of integration.)
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(Where C is a constant of integration.)
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Verified Answer
The correct answer is:
$\frac{x}{\log x}+C$
$\begin{aligned} & \int\{f(x)-g(x)\} d x=\int f(x) d x-\int g(x) d x \\ & =\int \frac{1}{\log x} d x-\int \frac{1}{(\log x)^2} d x \\ & =\int \frac{1}{\log x} \times 1 d x-\int \frac{1}{(\log x)^2} d x \\ & =\frac{1}{\log x} \cdot x-\int \frac{-1}{(\log x)^2} \times \frac{1}{x} \times x d x-\int \frac{1}{(\log x)^2} d x \\ & =\frac{x}{\log x}+\int \frac{1}{(\log x)^2} d x-\int \frac{d x}{(\log x)^2} \\ & =\frac{x}{\log x}\end{aligned}$
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