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Question: \(\int_{}^{}\left\{ \log(\log x) + \frac{1}{(\log x)^{2}} \right\}\) dx =...

∫{log⁡(log⁡x)+1(log⁡x)2}\int_{}^{}\left\{ \log(\log x) + \frac{1}{(\log x)^{2}} \right\} dx =

A

log (log x) + c

B

x log (log x) + cq

C

x{log⁡(log⁡x)−1log⁡x}\left\{ \log(\log x) - \frac{1}{\log x} \right\} + c

D

xlog⁡x\frac{x}{\log x} + c

Answer

x{log⁡(log⁡x)−1log⁡x}\left\{ \log(\log x) - \frac{1}{\log x} \right\} + c

Explanation

Solution

Put log x = t, x = et, dx = et dt

̃ ∫et(log⁡t+1t2)dt\int \mathrm { e } ^ { \mathrm { t } } \left( \log \mathrm { t } + \frac { 1 } { \mathrm { t } ^ { 2 } } \right) \mathrm { dt }

̃∫et{log⁡t−1t+1t+1t2}\int \mathrm { e } ^ { \mathrm { t } } \left\{ \log \mathrm { t } - \frac { 1 } { \mathrm { t } } + \frac { 1 } { \mathrm { t } } + \frac { 1 } { \mathrm { t } ^ { 2 } } \right\}dt [et (f(t) + f ' (t))]

= et (log⁡t−1t)\left( \log t - \frac{1}{t} \right)+ c = x{log⁡(log⁡x)−1log⁡x}x\left\{ \log(\log x) - \frac{1}{\log x} \right\} + c