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Question Number 189975 by mathlove last updated on 25/Mar/23

Answered by a.lgnaoui last updated on 27/Mar/23

$$Q_1\frac{lnx}{2-x}=-\frac{lnx}{x(1-\frac{2}{x})}=\frac{lnx}{x}\times \left(\frac{x}{2-x}\right)$$$$U^{{}^{\prime }}=\frac{lnx}{x}\to U=\frac{1}{2}{\left(lnx\right)}^2$$$$V=\frac{x}{2-x}\to V^{\prime }=\frac{2}{{(2-x)}^2}$$$$UV=\frac{x{\left(lnx\right)}^2}{2(2-x)}{UV}^{{}^{\prime }}=\frac{{\left(lnx\right)}^2}{{(x-2)}^2}$$$$=\frac{x^2}{{(x-2)}^2}\times {\left(\frac{lnx}{x}\right)}^2=\frac{{\left(lnx\right)}^2}{{(x-2)}^2}$$$$I=UV-\int {UV}^{{}^{\prime }}$$$$=\frac{x{\left(lnx\right)}^2}{2(x-2)}-\int \frac{{\left[ln((x-2)+2)\right]}^2}{{(x-2)}^2}$$$$$$$$x-2=tdx=dt$$$$-\int \frac{{\left[ln(t+2)\right]}^2}{t^2}dt$$$$u^{{}^{\prime }}=-\frac{1}{t^2}\to u=\frac{1}{t}$$$$v={\left[ln(t+2)\right]}^2\to v^{\prime }=2ln(t+2)\frac{1}{t+2}$$$$uv=\frac{ln(t+2)}{t}{uv}^{\prime }=\frac{2ln(t+2)}{t}$$$$=2\frac{lnx}{x-2}$$$$I=\frac{x{\left(lnx\right)}^2}{2(x-2)}+\frac{lnx}{x-2}-2\int \frac{lnx}{x-2}$$$$3I={\left[\frac{x{\left(lnx\right)}^2}{2(x-2)}\right]}_{\frac{3}{2}}^1+{\left[\frac{lnx}{x-2}\right]}_{\frac{3}{2}}^1$$$$I=\frac{1}{3}[(-\frac{\frac{3}{2}{(ln3-ln2)}^2}{2(\frac{3}{2}-2)})-\frac{ln3-ln2}{\frac{3}{2}-2}]$$$$=\frac{1}{3}\left(\frac{3(}{}\frac{{ln3-ln2)}^2}{2}\right)+2\times \frac{ln3-ln2}{3}$$$$=\frac{{(ln3-ln2)}^2}{2}+\frac{2(ln3-ln2)}{3}$$$$I=\frac{3{\left(ln3\right)}^2+3{\left(ln2\right)}^2-4(ln3-ln2)}{6}$$$$$$

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