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Question Number 83842 by M±th+et£s last updated on 06/Mar/20

$$\int \frac{ln\left(x\right)}{ln(6x-x^2)}dx$$

Commented by Henri Boucatchou last updated on 06/Mar/20

$$Using\frac{lnA}{lnB}=\frac{A}{B},\int \frac{x}{6x-x^2}dx=\int (-\frac{1}{2}\frac{2x}{6-x^2})dx=-\frac{1}{2}ln\mid 6-x^2\mid +Cte$$

Commented by MJS last updated on 06/Mar/20

$$\mathrm{but}\frac{\log A}{\log B}\ne \frac{A}{B}$$$$\frac{\log A}{\log B}=\frac{A}{B}\Rightarrow B\log A=A\log B\Rightarrow A^B=B^A$$$$\mathrm{which}\mathrm{obviously}\mathrm{is}\mathrm{not}\mathrm{generally}\mathrm{true}$$

Commented by M±th+et£s last updated on 06/Mar/20

$$thankyousir$$

Commented by MJS last updated on 07/Mar/20

$$\mathrm{I}\mathrm{tried}\mathrm{some}\mathrm{things}\mathrm{I}\mathrm{know}\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{possible}$$$$\mathrm{for}\mathrm{me}\mathrm{to}\mathrm{solve}\mathrm{it}$$

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