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Question Number 63883 by mmkkmm000m last updated on 10/Jul/19

$$\int ln\left(x\right)ln(1-x)ln(1-2x)dx$$

Commented by mathmax by abdo last updated on 12/Jul/19

$$letA=\int ln\left(x\right)ln(1-x)ln(1-2x)dxwehave$$$${ln}^{{}^{\prime }}(1-u)=-\frac{1}{1-u}=-\sum _{n=0}^{\mathrm{\infty }}{u}^n\Rightarrow ln(1-u)=-\sum _{n=0}^{\mathrm{\infty }}\frac{u^{n+1}}{n+1}+c(c=0)$$$$=-\sum _{n=1}^{\mathrm{\infty }}\frac{u^n}{n}(\mid u\mid <1)\Rightarrow for\mid x\mid <\frac{1}{2}weget$$$$ln(1-2x)=-\sum _{n=1}^{\mathrm{\infty }}\frac{2^n{x}^n}{n}andln(1-x)=-\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}\Rightarrow$$$$ln(1-x)ln(1-2x)=\left(\sum _{n=1}^{\mathrm{\infty }}\frac{2^n}{n}{x}^n\right).\left(\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}\right)$$$$=\sum _{n=1}^{\mathrm{\infty }}{c}_n{x}^{n}with{c}_n=\sum _{i+j=n}{a}_i{b}_j=\sum _{i=1}^n{a}_i{b}_{n-i}$$$$=\sum _{i=1}^n\frac{2^i}{i}\frac{1}{n-i}\Rightarrow ln(1-x)ln(1-2x)=\sum _{n=1}^{\mathrm{\infty }}\left(\sum _{i=1}^n\frac{2^i}{i(n-i)}\right)x^n\Rightarrow$$$$\int ln\left(x\right)ln(1-x)ln(1-2x)dx=\int lnx\left(\sum _{n=1}^{\mathrm{\infty }}\sum _{i=1}^n\frac{2^i}{i(n-i)}\right)x^{n}dx$$$$=\sum _{n=1}^{\mathrm{\infty }}\sum _{i=1}^n\frac{2^i}{i(n-i)}\int x^{n}ln\left(x\right)dx+c$$$$W_n=\int x^{n}ln\left(x\right)dxbyparts$$$$W_n=\frac{x^{n+1}}{n+1}ln\left(x\right)-\int \frac{x^{n+1}}{n+1}\frac{dx}{x}=\frac{x^{n+1}}{n+1}ln\left(x\right)-\frac{1}{(n+1)}\int x^{n}dx$$$$=\frac{x^{n+1}ln\left(x\right)}{n+1}-\frac{x^{n+1}}{{(n+1)}^2}\Rightarrow$$$$\int ln\left(x\right)ln(1-x)ln(1-2x)dx$$$$=\sum _{n=1}^{\mathrm{\infty }}\sum _{i=1}^n\frac{2^i}{i(n-i)}\frac{x^{n+1}ln\left(x\right)}{n+1}-\sum _{n=1}^{\mathrm{\infty }}\sum _{i=1}^n\frac{2^i}{i(n-i){(n+1)}^2}{x}^{n+1}+c$$$$$$

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