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Question Number 164547 by mnjuly1970 last updated on 21/Jan/22

prove Ω= ∫_0 ^( ∞) (( (√x))/(( 1+x +x^( 2) )^( 3) )) dx =^? ((π(√3))/(36)) −−m.n−−

proveΩ=∫0∞x(1+x+x2)3dx=?π336−−m.n−−

Answered by MJS_new last updated on 19/Jan/22

∫((√x)/((x^2 +x+2)^3 ))dx= [t=(√x) → dx=2(√x)dt] =2∫(t^2 /((t^4 +t^2 +1)^3 ))dt= [Ostrogradski′s Method] =((t(7t^6 +10t^4 +14t^2 +5))/(12(t^4 +t^2 +1)^2 ))+(1/(12))∫((7t^2 −5)/(t^4 +t^2 +1))dt= (1/(12))∫((7t^2 −5)/(t^4 +t^2 +1))dt= =(1/4)∫((2t−1)/(t^2 −t+1))dt+(1/(24))∫(dt/(t^2 −t+1))−(1/4)∫((2t+1)/(t^2 +t+1))dt+(1/(24))∫(dt/(t^2 +t+1))= ... =(1/4)ln ((t^2 −t+1)/(t^2 +t+1)) +((√3)/(36))(arctan (((√3)(2t−1))/3) +arctan (((√3)(2t+1))/3)) =((t(7t^6 +10t^4 +14t^2 +5))/(12(t^4 +t^2 +1)^2 ))+(1/4)ln ((t^2 −t+1)/(t^2 +t+1)) +((√3)/(36))(arctan (((√3)(2t−1))/3) +arctan (((√3)(2t+1))/3))= =(((√x)(7x^3 +10x^2 +14x+5))/(12(x^2 +x+1)^2 ))+(1/4)ln ((x−(√x)+1)/(x+(√x)+1)) +((√3)/(36))(arctan (((√3)(2(√x)−1))/3) +arctan (((√3)(2(√x)+1))/3))+C ⇒ ∫_0 ^∞ ((√x)/((x^2 +x+2)^3 ))dx=(((√3)π)/(36))

∫x(x2+x+2)3dx=[t=x→dx=2xdt]=2∫t2(t4+t2+1)3dt=[Ostrogradski′sMethod]=t(7t6+10t4+14t2+5)12(t4+t2+1)2+112∫7t2−5t4+t2+1dt=112∫7t2−5t4+t2+1dt==14∫2t−1t2−t+1dt+124∫dtt2−t+1−14∫2t+1t2+t+1dt+124∫dtt2+t+1=...=14lnt2−t+1t2+t+1+336(arctan3(2t−1)3+arctan3(2t+1)3)=t(7t6+10t4+14t2+5)12(t4+t2+1)2+14lnt2−t+1t2+t+1+336(arctan3(2t−1)3+arctan3(2t+1)3)==x(7x3+10x2+14x+5)12(x2+x+1)2+14lnx−x+1x+x+1+336(arctan3(2x−1)3+arctan3(2x+1)3)+C⇒∫∞0x(x2+x+2)3dx=3π36

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