int {left[ {frac{1}{{log x}} - frac{1}{{{{(lo | vedclass

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Question

(B) Let $I = \int {\left[ {\frac{1}{{\log x}} - \frac{1}{{{{(\log x)}^2}}}} \right]} \,dx$. Using integration by parts for the first term $\int \frac{

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int {\left[ {\frac{1}{{\log x}} - \frac{1}{{{{(\log x)}^2}}}} \right]} \,dx$. Using integration by parts for the first term $\int \frac{1}{\log x} dx$: Let $u = \frac{1}{\log x}$ and $dv = dx$. Then $du = -\frac{1}{x(\log x)^2} dx$ and $v = x$. So,$\int \frac{1}{\log x} dx = \frac{x}{\log x} - \int x \left( -\frac{1}{x(\log x)^2} \right) dx = \frac{x}{\log x} + \int \frac{1}{(\log x)^2} dx$. Substituting this back into the original integral: $I = \left( \frac{x}{\log x} + \int \frac{1}{(\log x)^2} dx \right) - \int \frac{1}{(\log x)^2} dx + c$. $I = \frac{x}{\log x} + c$.

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

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