Question

indegrate w.r.t X, 1/[xlogxlog(logx)]​

Answer 1

SOLUTION

TO DETERMINE

$\displaystyle \sf{ \int\limits_{}^{} \: \frac{1}{x logx log( logx ) } \: \, dx }$

EVALUATION

We have to find the value of

$\displaystyle \sf{ \int\limits_{}^{} \: \frac{1}{x logx log( logx ) } \: \, dx }$

$\displaystyle \sf{Let \: \: z \: = log( logx ) }$

Differentiating both sides with respect to x we get

$\displaystyle \sf{ \frac{dz}{dx} = \frac{d}{dx} \: log( logx ) }$

$\displaystyle \sf{ \implies \: \frac{dz}{dx} = \frac{1}{ logx} \: \frac{d}{dx} \: ( logx ) }$

$\displaystyle \sf{ \implies \: \frac{dz}{dx} = \frac{1}{ logx} \: \frac{1}{x} \: }$

$\displaystyle \sf{ \implies \: \frac{dz}{dx} = \frac{1}{x logx} \: }$

$\displaystyle \sf{ \implies \: dz = \frac{1}{x logx} dx \: }$

Now

$\displaystyle \sf{ \int\limits_{}^{} \: \frac{1}{x logx log( logx ) } \: \, dx }$

$\displaystyle \sf{ = \int\limits_{}^{} \: \frac{1}{z } \: \, dz }$

$\displaystyle \sf{ = log \: z + c }$

$\displaystyle \sf{ = log (log( logx )) + c }$

Where C is integration constant

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