`int{log(logx)+(1)/((logx)^(2))}dx=x {f (x)-g(x)}+C`, then

Options

A. `f(x)= log (log x), g(x) =(1)/(logx)`
B. `f(x)=log x, g(x)=(1)/(log x)`
C. `f(x)=(1)/(log x), g(x)= log (log x)`
D. `f(x)=(1)/(x log x), g (x) (1)/(log x)`

Answer

A

Explanation

Given, `int[log(logx)+(1)/((logx)^(2))]dx`
`int[log(logx)+(1)/((logx)^(2))]dx`
LHS
`=intunderset(||)1.log(logunderset(|)x)dx+int(1)/(logx)dx`
`+int(1)/((logx)^(2))dx`
`=xlog(logx)-(x)/(logx)int(1)/((logx)^(2))dx`
`+int(1)/((logx)^(2))dx+c`
`=x[log(logx)-(1)/(log)]+c`
`:.f(x)=log(logx),g(x)=(1)/(logx)`