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  4. if [log(logx) + (1/(logx)2)]dx = x[f(x)-g(x)] +C, then

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

Integrals

Solution:

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 = \int1\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 $

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