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Question Number 211151 by Spillover last updated on 29/Aug/24

$$$$$$$$$$\int \frac{dx}{x\ln x(\ln \ln x-1)(\ln \ln x-2)(\ln \ln x-3)(\ln \ln x-4)}$$$$$$$$$$

Answered by Ghisom last updated on 30/Aug/24

$$\mathrm{let}t=\ln \ln x\to dxx\ln xdt$$$$\int \frac{dt}{\underset{j=1}{\prod ^4}(t-j)}=$$$$...$$$$=-\frac{1}{6}\ln (t-1)+\frac{1}{2}\ln (t-2)-\frac{1}{2}\ln (t-3)+\frac{1}{6}\ln (t-4)=$$$$=\frac{1}{6}\ln \mid \frac{{(t-2)}^3(t-4)}{(t-1){(t-3)}^3}\mid =...$$

Commented by Spillover last updated on 31/Aug/24

$$good$$

Answered by Spillover last updated on 31/Aug/24

$$\int \frac{dx}{x\ln x(\ln \ln x-1)(\ln \ln x-2)(\ln \ln x-3)(\ln \ln x-4)}$$$$t=\ln \ln xdt=\frac{dx}{x\ln x}$$$$\int \frac{dt}{(t-1)(t-2)(t-3)(t-4)}$$$$$$$$\frac{1}{3}\int \frac{(t-2)(t-3)-(t-1)(t-4)}{(t-1)(t-2)(t-3)(t-4)}dt$$$$$$$$\frac{1}{3}[\int \frac{dt}{(t-1)(t-4)}.\frac{1}{2}]-[\int \frac{1}{2}.\frac{dt}{(t-2)(t-3)}]$$$$$$$$\frac{1}{6}\int \frac{(t-1)-(t-4)}{(t-1)(t-4)}-\frac{1}{2}[\int (\frac{1}{t-3}-\frac{1}{t-2})dt]$$$$$$$$\frac{1}{6}\ln \left(\frac{t-4}{t-1}\right)+\frac{1}{2}\ln \left(\frac{t-2}{t-3}\right)$$$$$$$$butt=\ln \ln x$$$$$$$$\frac{1}{6}\ln \left(\frac{\ln \ln x-4}{\ln \ln x-1}\right)+\frac{1}{2}\ln \left(\frac{\ln \ln x-2}{\ln \ln x-3}\right)$$

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