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

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

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$x \left[ \log(\log x) - \frac{1}{\log x} \right] + c$

Vedclass · Answer

(D) Let $I = \int \left[ \log(\log x) + \frac{1}{(\log x)^2} \right] dx$. Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$. Then,$I = \int e^t \left( \log t + \frac{1}{t^2} \right) dt$. We can rewrite the integrand as: $I = \int e^t \left[ \left( \log t + \frac{1}{t} \right) - \left( \frac{1}{t} - \frac{1}{t^2} \right) \right] dt$. Using the standard integral formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + C$,where $f(t) = \log t$ and $f'(t) = \frac{1}{t}$,we get: $I = e^t \log t - \int e^t \left( \frac{1}{t} - \frac{1}{t^2} \right) dt$. Since $\frac{d}{dt} \left( \frac{1}{t} \right) = -\frac{1}{t^2}$,the second integral is $\int e^t \left( \frac{1}{t} + \frac{d}{dt} \left( \frac{1}{t} \right) \right) dt = e^t \left( \frac{1}{t} \right) + C$. Thus,$I = e^t \log t - e^t \left( \frac{1}{t} \right) + C$. Substituting $t = \log x$ and $e^t = x$ back: $I = x \log(\log x) - \frac{x}{\log x} + C = x \left[ \log(\log x) - \frac{1}{\log x} \right] + C$.

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

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