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Question-149241




Question Number 149241 by john_santu last updated on 04/Aug/21
Answered by ajfour last updated on 04/Aug/21
s=∫_0 ^( 1) ((x^3 dx)/((x−1)^3 +(3−h)(x−1)+hx))  3−h=5   ⇒  h=−2  s_1 =∫((x^3 dx)/((x−1)^3 +5(x−1)−2x))  s_2 =∫(((1−x)^3 dx)/(−x^3 −5x−2(1−x)))      =∫(((x−1)^3 dx)/(x^3 +5x−2(x−1)))    =∫(((x−1)^3 dx)/(x^3 +3x+2))  trying factors of d^r   one is  p=(((√2)−1))^(1/3) −(((√2)+1))^(1/3)    s =∫_0 ^( 1) (((x−1)^3 dx)/((x−p)(x^2 +qx+r)))  we′ll employ partial fractions.
$${s}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{x}^{\mathrm{3}} {dx}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} +\left(\mathrm{3}−{h}\right)\left({x}−\mathrm{1}\right)+{hx}} \\ $$$$\mathrm{3}−{h}=\mathrm{5}\:\:\:\Rightarrow\:\:{h}=−\mathrm{2} \\ $$$${s}_{\mathrm{1}} =\int\frac{{x}^{\mathrm{3}} {dx}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} +\mathrm{5}\left({x}−\mathrm{1}\right)−\mathrm{2}{x}} \\ $$$${s}_{\mathrm{2}} =\int\frac{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} {dx}}{−{x}^{\mathrm{3}} −\mathrm{5}{x}−\mathrm{2}\left(\mathrm{1}−{x}\right)} \\ $$$$\:\:\:\:=\int\frac{\left({x}−\mathrm{1}\right)^{\mathrm{3}} {dx}}{{x}^{\mathrm{3}} +\mathrm{5}{x}−\mathrm{2}\left({x}−\mathrm{1}\right)} \\ $$$$\:\:=\int\frac{\left({x}−\mathrm{1}\right)^{\mathrm{3}} {dx}}{{x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{2}} \\ $$$${trying}\:{factors}\:{of}\:{d}^{{r}} \\ $$$${one}\:{is}\:\:{p}=\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{2}}−\mathrm{1}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{2}}+\mathrm{1}} \\ $$$$\:{s}\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\left({x}−\mathrm{1}\right)^{\mathrm{3}} {dx}}{\left({x}−{p}\right)\left({x}^{\mathrm{2}} +{qx}+{r}\right)} \\ $$$${we}'{ll}\:{employ}\:{partial}\:{fractions}. \\ $$
Answered by john_santu last updated on 04/Aug/21

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