Question Number 214915 by Spillover last updated on 23/Dec/24
Answered by maths2 last updated on 23/Dec/24
$${x}\rightarrow\mathrm{1}−{x};{Let}\:{A}\:{bee}\:{the}\:{integral} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left(\mathrm{1}+{e}^{−\mathrm{2}+\mathrm{4}{x}} \right)\left(\mathrm{5}+\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} \right)}\Rightarrow \\ $$$$\mathrm{2}{A}=\int_{\mathrm{0}} ^{\mathrm{1}} \left[\frac{\mathrm{1}}{\left(\mathrm{5}+\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} \right)}.\left(\frac{\mathrm{1}}{\mathrm{1}+{e}^{−\mathrm{2}+\mathrm{4}{x}} }+\frac{\mathrm{1}}{\mathrm{1}+{e}^{\mathrm{2}−\mathrm{4}{x}} }\right){dx}\right. \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{5}+\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} }..{easy}\:{from}\:{Hear} \\ $$