Question

Integrate the following with respect to x

$\frac{1}{xlogxlog(logx)}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int \frac{1}{x \: logx \: log(logx)} \: dx$

To solve this integral, we use Method of Substitution.

So, Substitute

$\red{\rm :\longmapsto\:log(logx) = y}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx} log(logx) = \dfrac{d}{dx} y}$

$\red{\rm :\longmapsto\:\dfrac{1}{x \: logx} \: dx \: = \: dy}$

So, on substituting the values, we get

$\rm \:  =  \: \displaystyle\int \frac{1}{y} \: dy$

$\rm \:  =  \: log |y| + c$

$\rm \:  =  \: log |log \: (logx)| + c$

Hence,

$\rm \implies\:\boxed{\tt{ \displaystyle\int \frac{1}{x \: logx log(logx)}dx = log |log \: (logx)| + c}}$

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Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$