Question
$$I = \int {\left\{ {{{\log }_e}{{\log }_e}x + \frac{1}{{{{\left( {{{\log }_e}x} \right)}^2}}}} \right\}} dx$$ is equal to :
A.
$$x\,{\log _e}{\log _e}x + c$$
B.
$$x\,{\log _e}{\log _e}x - \frac{x}{{{{\log }_e}x}} + c$$
C.
$$x\,{\log _e}{\log _e}x + \frac{x}{{{{\log }_e}x}} + c$$
D.
none of these
Answer :
$$x\,{\log _e}{\log _e}x - \frac{x}{{{{\log }_e}x}} + c$$
Solution :
$$\eqalign{
& {\text{Put }}ln\,x = t \cr
& \Rightarrow \frac{1}{x}dx = dt \Rightarrow dx = x\,dt = {e^t}dt \cr
& \therefore \,I = \int {\left( {ln\,t + \frac{1}{{{t^2}}}} \right){e^t}dt} \cr
& = \int {\left( {ln\,t + \frac{1}{t} - \frac{1}{t} + \frac{1}{{{t^2}}}} \right){e^t}dt} \cr
& = \int {\left( {ln\,t + \frac{1}{t}} \right){e^t}dt + } \int {{e^t}\left( { - \frac{1}{t} + \frac{1}{{{t^2}}}} \right)dt} \cr
& = {e^t}ln\,t - \frac{{{e^t}}}{t} + c\,\,\,\,\,\left[ {\because \,\frac{d}{{dt}}ln\,t = \frac{1}{t}{\text{ and }}\frac{d}{{dt}}\left( { - \frac{1}{t}} \right) = \frac{1}{{{t^2}}}} \right] \cr
& = x\,ln\left( {ln\,x} \right) - \frac{x}{{ln\,x}} + c \cr} $$