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.
Correct Answer: B
Solution :
[b] Put, \[l\,nx=t\] \[\Rightarrow \frac{1}{x}dx=dt\Rightarrow dx=xdt={{e}^{t}}dt\] \[\therefore I=\int{\left( l\,n\,\,t+\frac{1}{{{t}^{2}}} \right){{e}^{t}}dt}\] \[=\int{\left( l\,n\,\,t+\frac{1}{t}-\frac{1}{t}+\frac{1}{{{t}^{2}}} \right){{e}^{t}}dt}\] \[=\int{\left( l\,n\,\,t+\frac{1}{t} \right){{e}^{t}}dt}+\int{{{e}^{t}}\left( -\frac{1}{t}+\frac{1}{{{t}^{2}}} \right)dt}\] \[={{e}^{t}}ln\,t-\frac{{{e}^{t}}}{t}+c\left[ \because \frac{d}{dt}l\,n\,\,t=\frac{1}{t}and\,\frac{d}{dt}\,\left( -\frac{1}{t} \right)=\frac{1}{{{t}^{2}}} \right]\] \[=x\,l\,n(l\,n\,\,x)-\frac{x}{l\,n\,x}+c\]
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