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ln-x-ln-6x-x-2-dx-




Question Number 83842 by M±th+et£s last updated on 06/Mar/20
∫((ln(x))/(ln(6x−x^2 )))dx
$$\int\frac{{ln}\left({x}\right)}{{ln}\left(\mathrm{6}{x}−{x}^{\mathrm{2}} \right)}{dx} \\ $$
Commented by Henri Boucatchou last updated on 06/Mar/20
Using   ((lnA)/(lnB))=(A/B),        ∫(x/(6x−x^2 ))dx=∫(−(1/2)((2x)/(6−x^2 )))dx=−(1/2)ln∣6−x^2 ∣ + Cte
$${Using}\:\:\:\frac{{lnA}}{{lnB}}=\frac{{A}}{{B}},\:\:\:\:\:\:\:\:\int\frac{{x}}{\mathrm{6}{x}−{x}^{\mathrm{2}} }{dx}=\int\left(−\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{2}{x}}{\mathrm{6}−{x}^{\mathrm{2}} }\right){dx}=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\mathrm{6}−{x}^{\mathrm{2}} \mid\:+\:{Cte} \\ $$
Commented by MJS last updated on 06/Mar/20
but ((log A)/(log B))≠(A/B)  ((log A)/(log B))=(A/B) ⇒ Blog A =Alog B ⇒ A^B =B^A   which obviously is not generally true
$$\mathrm{but}\:\frac{\mathrm{log}\:{A}}{\mathrm{log}\:{B}}\neq\frac{{A}}{{B}} \\ $$$$\frac{\mathrm{log}\:{A}}{\mathrm{log}\:{B}}=\frac{{A}}{{B}}\:\Rightarrow\:{B}\mathrm{log}\:{A}\:={A}\mathrm{log}\:{B}\:\Rightarrow\:{A}^{{B}} ={B}^{{A}} \\ $$$$\mathrm{which}\:\mathrm{obviously}\:\mathrm{is}\:\mathrm{not}\:\mathrm{generally}\:\mathrm{true} \\ $$
Commented by M±th+et£s last updated on 06/Mar/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by MJS last updated on 07/Mar/20
I tried some things I know but it′s not possible  for me to solve it
$$\mathrm{I}\:\mathrm{tried}\:\mathrm{some}\:\mathrm{things}\:\mathrm{I}\:\mathrm{know}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{possible} \\ $$$$\mathrm{for}\:\mathrm{me}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it} \\ $$

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