If $\mathrm{f}(\mathrm{x})=\frac{1}{\log \mathrm{x}}, \mathrm{g}(\mathrm{x})=\frac{1}{(\log \mathrm{x})^2}$, then find $\int\{f(x)-g(x)\} d x$ (Where C is a constant of integration.)

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Aicharya · Accepted Answer

$\begin{aligned} & \int\{f(x)-g(x)\} d x=\int f(x) d x-\int g(x) d x \ & =\int \frac{1}{\log x} d x-\int \frac{1}{(\log x)^2} d x \ & =\int \frac{1}{\log x} 	imes 1 d x-\int \frac{1}{(\log x)^2} d x \ & =\frac{1}{\log x} \cdot x-\int \frac{-1}{(\log x)^2} 	imes \frac{1}{x} 	imes x d x-\int \frac{1}{(\log x)^2} d x \ & =\frac{x}{\log x}+\int \frac{1}{(\log x)^2} d x-\int \frac{d x}{(\log x)^2} \ & =\frac{x}{\log x}\end{aligned}$

If $\mathrm{f}(\mathrm{x})=\frac{1}{\log \mathrm{x}}, \mathrm{g}(\mathrm{x})=\frac{1}{(\log \mathrm{x})^2}$,…

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