int {left[ {log (log x) + frac{1}{{{{(log x)} | vedclass

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Question

(B) Let $I = \int {\left[ {\log (\log x) + \frac{1}{{{{(\log x)}^2}}}} \right]\,dx}$.We can split the integral as $I = \int {\log (\log x)\,dx + \int

Vedclass · Accepted Answer

$x\log (\log x) - \frac{x}{{\log x}} + c$

Vedclass · Answer

(B) Let $I = \int {\left[ {\log (\log x) + \frac{1}{{{{(\log x)}^2}}}} \right]\,dx}$.We can split the integral as $I = \int {\log (\log x)\,dx + \int {\frac{1}{{{{(\log x)}^2}}}} \,dx}$.Applying integration by parts to the first term $\int {\log (\log x)\,dx}$,let $u = \log (\log x)$ and $dv = dx$. Then $du = \frac{1}{\log x} \cdot \frac{1}{x} dx$ and $v = x$.So,$\int {\log (\log x)\,dx} = x \log (\log x) - \int {x \cdot \frac{1}{x \log x} \,dx} = x \log (\log x) - \int {\frac{1}{\log x} \,dx}$.Now,apply integration by parts to $\int {\frac{1}{\log x} \,dx}$. Let $u = \frac{1}{\log x}$ and $dv = dx$. Then $du = -\frac{1}{(\log x)^2} \cdot \frac{1}{x} dx$ and $v = x$.Thus,$\int {\frac{1}{\log x} \,dx} = \frac{x}{\log x} - \int {x \cdot \left( -\frac{1}{x(\log x)^2} \right) \,dx} = \frac{x}{\log x} + \int {\frac{1}{(\log x)^2} \,dx}$.Substituting this back into the expression for $I$:$I = x \log (\log x) - \left( \frac{x}{\log x} + \int {\frac{1}{(\log x)^2} \,dx} \right) + \int {\frac{1}{(\log x)^2} \,dx}$.$I = x \log (\log x) - \frac{x}{\log x} + c$.

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

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