Question

if $\left[log\left(logx\right)+\frac{1}{{\left(logx\right)}^2}\right]dx=x\left[f\left(x\right)-g\left(x\right)\right]+C,then$

• A. $f\left(x\right)=log\left(logx\right),g\left(x\right)=\frac{1}{logx}$
• B. $f\left(x\right)=logx,g\left(x\right)=\frac{1}{logx}$
• C. $f\left(x\right)=\frac{1}{logx},g\left(x\right)=\frac{1}{logx}$
• D. $f\left(x\right)=\frac{1}{xlogx},g\left(x\right)=\frac{1}{logx}$

Answer 1

Answer: (A)

Given,

$\int \left[log\left(logx\right)+\frac{1}{{\left(log\right)}^2}\right]dx=x\left[f\left(x\right)-g\left(x\right)\right]+C$

$LHS=\int 1\cdot log\left(logx\right)dx+\int \frac{1}{{\left(logx\right)}^2}dx$

$=log\left(logx\right)\cdot x-\int \left(\frac{1}{\left(logx\right)\times x}\times x\right)dx$

$+\int \frac{1}{{\left(logx\right)}^2}dx+C$

$=xlog\left(logx\right)-\int \frac{1}{logx}dx+\int \frac{1}{{\left(logx\right)}^2}dx+C$

$=xlog\left(logx\right)-\frac{1}{logx}\times x+\int \frac{-1}{{\left(logx\right)}^2}\cdot \frac{x}{x}dx+\int \frac{1}{{\left(logx\right)}^2}dx+C$

$=x\left[log\left(logx\right)-\frac{1}{logx}\right]+C$

$\therefore f\left(x\right)=log\left(logx\right),g\left(x\right)=\frac{1}{logx}$