Question Number 149241 by john_santu last updated on 04/Aug/21
Answered by ajfour last updated on 04/Aug/21
$${s}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{x}^{\mathrm{3}} {dx}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} +\left(\mathrm{3}−{h}\right)\left({x}−\mathrm{1}\right)+{hx}} \\ $$$$\mathrm{3}−{h}=\mathrm{5}\:\:\:\Rightarrow\:\:{h}=−\mathrm{2} \\ $$$${s}_{\mathrm{1}} =\int\frac{{x}^{\mathrm{3}} {dx}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} +\mathrm{5}\left({x}−\mathrm{1}\right)−\mathrm{2}{x}} \\ $$$${s}_{\mathrm{2}} =\int\frac{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} {dx}}{−{x}^{\mathrm{3}} −\mathrm{5}{x}−\mathrm{2}\left(\mathrm{1}−{x}\right)} \\ $$$$\:\:\:\:=\int\frac{\left({x}−\mathrm{1}\right)^{\mathrm{3}} {dx}}{{x}^{\mathrm{3}} +\mathrm{5}{x}−\mathrm{2}\left({x}−\mathrm{1}\right)} \\ $$$$\:\:=\int\frac{\left({x}−\mathrm{1}\right)^{\mathrm{3}} {dx}}{{x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{2}} \\ $$$${trying}\:{factors}\:{of}\:{d}^{{r}} \\ $$$${one}\:{is}\:\:{p}=\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{2}}−\mathrm{1}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{2}}+\mathrm{1}} \\ $$$$\:{s}\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\left({x}−\mathrm{1}\right)^{\mathrm{3}} {dx}}{\left({x}−{p}\right)\left({x}^{\mathrm{2}} +{qx}+{r}\right)} \\ $$$${we}'{ll}\:{employ}\:{partial}\:{fractions}. \\ $$
Answered by john_santu last updated on 04/Aug/21
