If $\int\left\{\log (\log x)+\frac{1}{(\log x)^2}\right\} d x=x\{f(x)-g(x)\}+C,$ then

$f(x)=\log x ; g(x)=\frac{1}{\log x}$

$f(x)=\frac{1}{\log x} ;(x)=\log (\log x)$

$f(x)=\frac{1}{x \log x} ; g(x)=\frac{1}{\log x}$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

\int\left\{\log (\log x)+\frac{1}{(\log x)^2}\right\} d x=x\{f(x)-g(x)\}+C,

then

View answer and explanation

Correct option: A

f(x)=\log (\log x) ; g(x)=\frac{1}{\log x}

Option A (

f(x)=\log (\log x) ; g(x)=\frac{1}{\log x}

) is correct because after integration by parts and substitution, the given integral matches the form

x\{f(x)-g(x)\} + C