int left[ frac{1}{log x} - frac{1}{(log x)^2} | vedclass

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Question

(D) Let $I = \int \left( \frac{1}{\log x} - \frac{1}{(\log x)^2} \right) dx$. Substitute $\log x = t$,which implies $x = e^t$. Then,$dx = e^t dt$. Sub

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $I = \int \left( \frac{1}{\log x} - \frac{1}{(\log x)^2} \right) dx$. Substitute $\log x = t$,which implies $x = e^t$. Then,$dx = e^t dt$. Substituting these into the integral,we get: $I = \int \left( \frac{1}{t} - \frac{1}{t^2} \right) e^t dt$. We know the standard integration formula: $\int e^t (f(t) + f'(t)) dt = e^t f(t) + C$. Here,$f(t) = \frac{1}{t}$ and $f'(t) = -\frac{1}{t^2}$. Therefore,$I = e^t \cdot \frac{1}{t} + c$. Substituting back $t = \log x$ and $e^t = x$: $I = \frac{x}{\log x} + c$. Thus,option $D$ is correct.

$\int \left[ \frac{1}{\log x} - \frac{1}{(\log x)^2} \right] dx =$

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