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Question-208540




Question Number 208540 by otchereabdullai@gmail.com last updated on 17/Jun/24
Answered by Sutrisno last updated on 18/Jun/24
misal : 3x=2tanθ  x=(2/3)tanθ→dx=(2/3)sec^2 θdθ  =∫(((2/3)tanθ−3)/( (√((2tanθ)^2 +4)))).(2/3)sec^2 θdθ  =∫(((2/3)tanθ−3)/( (√(4(tan^2 θ+1))))).(2/3)sec^2 θdθ  =∫(((2/3)tanθ−3)/( 2secθ)).(2/3)sec^2 θdθ  =∫(1/3)((2/3)tanθ−3)secθdθ  =∫(1/3)((2/3)tanθsecθ−3secθ)dθ  =(1/3)((2/3)secθ−3ln(secθ+tanθ))+c  =(1/3)((2/3)((√(9x^2 +4))/2)−3ln(((√(9x^2 +4))/2)+((3x)/2)))+c
misal:3x=2tanθx=23tanθdx=23sec2θdθ=23tanθ3(2tanθ)2+4.23sec2θdθ=23tanθ34(tan2θ+1).23sec2θdθ=23tanθ32secθ.23sec2θdθ=13(23tanθ3)secθdθ=13(23tanθsecθ3secθ)dθ=13(23secθ3ln(secθ+tanθ))+c=13(239x2+423ln(9x2+42+3x2))+c
Answered by Frix last updated on 18/Jun/24
∫((x−3)/( (√(9x^2 +4))))dx =^(t=((3x+(√(9x^3 +4)))/2))  ∫((1/9)−(1/t)−(1/(9t^2 )))dt=  =(t/9)−ln t +(1/(9t))=((t^2 +1)/(9t))−ln t =  =((√(9x^2 +4))/9)−ln (3x+(√(9x^2 +4))) +C
x39x2+4dx=t=3x+9x3+42(191t19t2)dt==t9lnt+19t=t2+19tlnt==9x2+49ln(3x+9x2+4)+C

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