Question

$\int_{}^{}\left\{ \log(\log x) + \frac{1}{(\log x)^{2}} \right\}$dx is equal to –

• A. x log (log x) – $\frac{x}{\log x}$+ c
• B. x log x – $\frac{x}{\log x}$ + c
• C. x^x log x – $\frac{x}{\log x}$+ c
• D. x log x – $\frac{x(\log x)}{\log x}$ + c

Answer 1

Answer: (A)

x log (log x) – $\frac{x}{\log x}$+ c

Answer 2

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

\ I = $\int \left\{ \log \mathrm { t } + \frac { 1 } { \mathrm { t } ^ { 2 } } \right\}$ e^t dt, where t = log x

= e^t dt

= $\int \left\{ \log t + \frac { 1 } { t } \right\}$ e^t dt + e^t dt

= $\int \mathrm { e } ^ { \mathrm { t } }$ log t dt + . dt + $\int \mathrm { e } ^ { \mathrm { t } }$ (–1/t) dt + $\int \mathrm { e } ^ { \mathrm { t } }$

II I II I

= (log t) e^t – . e^t dt + . dt + e^t

– . e^t dt + dt + c

= e^t . log t – e^t + c = x log (log x) – + c.

Hence (1) is the correct answer