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Question Number 156933 by CAIMAN last updated on 17/Oct/21

Answered by mindispower last updated on 17/Oct/21

$$\frac{1}{t}=x$$$${\int }_{\frac{1}{2}}^{\frac{1}{3}}\frac{ln\left(x\right)}{x(1+x)}dx=\int \frac{ln\left(x\right)}{x}-\frac{ln\left(x\right)}{1+x}dx$$$$=\frac{{ln}^2\left(x\right)}{2}-ln\left(x\right)ln(1+x)+\int \frac{ln(1+x)}{x}dx$$$$y=-xin\int \frac{ln(1+x)}{x}dx$$$$=\int \frac{ln(1-y)}{y}dy=-{Li}_2\left(y\right),{Li}_2...dilogarithmfunction$$$$\int \frac{ln\left(x\right)}{x(1+x)}=\frac{{ln}^2\left(x\right)}{2}-ln\left(x\right)ln(1+x)-{Li}_2(-x)+c$$$${\int }_{\frac{1}{2}}^{\frac{1}{3}}\frac{ln\left(x\right)}{x(1+x)}dx=\frac{1}{2}({ln}^2\left(3\right)-{ln}^2\left(3\right))-ln\left(\frac{1}{3}\right)ln\left(\frac{4}{3}\right)+ln\left(\frac{1}{2}\right)ln\left(\frac{3}{2}\right)$$$$-{li}_2(-\frac{1}{3})+{li}_2(-\frac{1}{2})$$$$$$$$$$

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