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prove-0-dx-x-4-25x-2-160-0-dx-x-4-95x-2-2560-




Question Number 212461 by MrGaster last updated on 14/Oct/24
                          prove:    ∫_0 ^(+∞) (dx/( (√(x^4 +25x^2 +160))))=∫_0 ^(+∞) (dx/( (√(x^4 −95x^2 +2560))))
prove:0+dxx4+25x2+160=0+dxx495x2+2560
Commented by Ghisom last updated on 19/Oct/24
do you know the solution?  I tried this but...    ∫_0 ^∞ (dx/( (√(x^4 +ax^2 +b))))=       [t=2arctan (x/( (b)^(1/4) )) → dx=((x^2 +(√b))/(2(b)^(1/4) ))dt]  =(√2)∫_0 ^(2π) (dt/( (√((6(√b)+a)+(2(√b)−a)cos 2t))))=  =2(√2)∫_0 ^π (dt/( (√((6(√b)+a)+(2(√b)−a)cos 2t))))=       [let c=((6(√b)+a)/(2(√b)−a))]  =((√(c+1))/( (b)^(1/4) ))∫_0 ^π (dt/( (√(c+cos 2t))))=(1/( (b)^(1/4) ))∫_0 ^π (dt/( (√(1−((2sin^2  t)/(c+1))))))=  =(1/( (b)^(1/4) ))[F (t∣(2/(c+1)))]_0 ^π =  =(1/( (b)^(1/4) ))[F (t∣(1/2)−(a/(4(√b))))]_0 ^π =  =(1/( (b)^(1/4) ))F (π∣(1/2)−(a/(4(√b))))    now the question is  (1/(2((10))^(1/4) ))F (π∣((16−5(√(10)))/(32)))  =^(???)  (1/(4((10))^(1/4) ))F (π∣((64+19(√(10)))/(128)))  ⇔  2F (π∣((16−5(√(10)))/(32)))  =^(???)  F (π∣((64+19(√(10)))/(128)))
doyouknowthesolution?Itriedthisbut0dxx4+ax2+b=[t=2arctanxb4dx=x2+b2b4dt]=22π0dt(6b+a)+(2ba)cos2t==22π0dt(6b+a)+(2ba)cos2t=[letc=6b+a2ba]=c+1b4π0dtc+cos2t=1b4π0dt12sin2tc+1==1b4[F(t2c+1)]0π==1b4[F(t12a4b)]0π==1b4F(π12a4b)nowthequestionis12104F(π1651032)=???14104F(π64+1910128)2F(π1651032)=???F(π64+1910128)

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