Integral Calculus Question 250
Question: $ I=\int{{ {\log_{e}}{\log_{e}}x+\frac{1}{{{({\log_{e}}x)}^{2}}} }dx} $ is equal to:
Options:
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.
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Put, $ l,nx=t $
$ \Rightarrow \frac{1}{x}dx=dt\Rightarrow dx=xdt=e^{t}dt $
$ \therefore I=\int{( l,nt+\frac{1}{t^{2}} )e^{t}dt} $ $ =\int{( l,nt+\frac{1}{t}-\frac{1}{t}+\frac{1}{t^{2}} )e^{t}dt} $ $ =\int{( l,nt+\frac{1}{t} )e^{t}dt}+\int{e^{t}( -\frac{1}{t}+\frac{1}{t^{2}} )dt} $ $ =e^{t}ln,t-\frac{e^{t}}{t}+c[ \because \frac{d}{dt}l,nt=\frac{1}{t}and,\frac{d}{dt},( -\frac{1}{t} )=\frac{1}{t^{2}} ] $ $ =x,l,n(l,nx)-\frac{x}{l,n,x}+c $
BETA
