If f(x) = frac{1}{log x} and g(x) = frac{1}{( | vedclass

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(C) We need to evaluate the integral $I = \int \left( \frac{1}{\log x} - \frac{1}{(\log x)^2} \right) dx$. Applying the linearity property of integral

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$\frac{x}{\log x} + C$

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(C) We need to evaluate the integral $I = \int \left( \frac{1}{\log x} - \frac{1}{(\log x)^2} \right) dx$. Applying the linearity property of integrals: $I = \int \frac{1}{\log x} dx - \int \frac{1}{(\log x)^2} dx$. Using integration by parts on the first term $\int \frac{1}{\log x} dx$,let $u = \frac{1}{\log x}$ and $dv = dx$. Then $du = -\frac{1}{(\log x)^2} \cdot \frac{1}{x} dx$ and $v = x$. Using the formula $\int u dv = uv - \int v du$: $\int \frac{1}{\log x} dx = \frac{x}{\log x} - \int x \left( -\frac{1}{(\log x)^2} \cdot \frac{1}{x} \right) dx$ $= \frac{x}{\log x} + \int \frac{1}{(\log x)^2} dx$. Substituting this back into the original expression for $I$: $I = \left( \frac{x}{\log x} + \int \frac{1}{(\log x)^2} dx \right) - \int \frac{1}{(\log x)^2} dx$ $I = \frac{x}{\log x} + C$.

If $f(x) = \frac{1}{\log x}$ and $g(x) = \frac{1}{(\log x)^2}$,then evaluate the integral $\int \{f(x) - g(x)\} dx$. (Where $C$ is a constant of integration.)

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