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

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(C) Let $I = \int \left( \frac{1}{\log x} - \frac{1}{(\log x)^2} \right) dx$.Consider the first term $\int \frac{1}{\log x} dx$. Let $u = \log x$, the

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

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(C) Let $I = \int \left( \frac{1}{\log x} - \frac{1}{(\log x)^2} \right) dx$.Consider the first term $\int \frac{1}{\log x} dx$. Let $u = \log x$, then $x = e^u$ and $dx = e^u du$.So, $\int \frac{1}{\log x} dx = \int \frac{e^u}{u} du$.Using integration by parts for $\int \frac{e^u}{u} du$ is not standard, but we can observe the derivative of $\frac{x}{\log x}$:$\frac{d}{dx} \left( \frac{x}{\log x} \right) = \frac{(\log x)(1) - x(\frac{1}{x})}{(\log x)^2} = \frac{\log x - 1}{(\log x)^2} = \frac{\log x}{(\log x)^2} - \frac{1}{(\log x)^2} = \frac{1}{\log x} - \frac{1}{(\log x)^2}$.Thus, $\int \left( \frac{1}{\log x} - \frac{1}{(\log x)^2} \right) dx = \frac{x}{\log x} + c$.

If $f(x) = \frac{1}{\log x}$ and $g(x) = \frac{1}{(\log x)^2}$, then the value of $\int [f(x) - g(x)] dx$ is...

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