यदि f(x) = frac{1}{log x} और g(x) = frac{1}{( | vedclass

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(C) माना $I = \int \left( \frac{1}{\log x} - \frac{1}{(\log x)^2} \right) dx$ है।हम $\frac{x}{\log x}$ का अवकलन ज्ञात करते हैं:$\frac{d}{dx} \left( \f

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

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(C) माना $I = \int \left( \frac{1}{\log x} - \frac{1}{(\log x)^2} \right) dx$ है।हम $\frac{x}{\log x}$ का अवकलन ज्ञात करते हैं:$\frac{d}{dx} \left( \frac{x}{\log x} \right) = \frac{(\log x)(1) - x(\frac{1}{x})}{(\log x)^2} = \frac{\log x - 1}{(\log x)^2} = \frac{\log x}{(\log x)^2} - \frac{1}{(\log x)^2} = \frac{1}{\log x} - \frac{1}{(\log x)^2}$।अतः, $\int \left( \frac{1}{\log x} - \frac{1}{(\log x)^2} \right) dx = \frac{x}{\log x} + c$।

यदि $f(x) = \frac{1}{\log x}$ और $g(x) = \frac{1}{(\log x)^2}$ है, तो $\int [f(x) - g(x)] dx$ का मान है...

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