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प्रश्न
find : `int(3x+1)sqrt(4-3x-2x^2)dx`
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उत्तर
`Let 3x+1=λd/dx(4−3x−2x^2)+μ`
⇒3x+1=λ(−3−4x)+μ
⇒3x+1=−3λ+μ−4λx
⇒3=−4λ , −3λ+μ=1
⇒λ=−3/4, μ=−5/4
`I=int(3x+1)sqrt(4-3x-2x^2)dx`
`=∫[−3/4(−3−4x)−5/4]sqrt(4−3x−2x^2)dx`
`=∫−3/4(−3−4x)sqrt(4−3x−2x^2)dx−∫5/4 sqrt(4−3x−2x^2)dx`
`=−3/4∫(−3−4x)sqrt(4−3x−2x^2)dx−5/4∫sqrt(4−3x−2x^2)dx `
Let 4−3x−2x2=t in the first integral⇒(−3−4x)dx=dt
`∴ I=−3/4∫sqrtt dt−5/4∫sqrt(−2(x^2+3/2x−2)dx`
`=−3/4×2/3t^(3/2)+C_1−5/4∫sqrt(−2(x^2+3/2x−2+9/16−9/16)dx`
`=−1/2(4−3x−2x^2)^(3/2)+C_1−5/4∫sqrt(−2[(x+3/4)^2−(sqrt41/4)^2])dx`
`=−1/2(4−3x−2x^2)^(3/2)+C_1−(5sqrt2)/4∫sqrt((sqrt41/4)^2−(x+3/4)^2)dx`
`=−1/2(4−3x−2x^2)^(3/2)+C_1-(5sqrt2)/4[1/2(x+3/4)sqrt((41/16)−(x+3/4)^2)+1/2(41/16)sin^−1 ((x+3/4)/(sqrt41/4))+C_2]`
`=−1/2(4−3x−2x^2)^(3/2)−5/(4sqrt2)(x+3/4)sqrt((41/16)−(x+3/4)^2)-205/(64sqrt2) sin^−1 ((4x+3)/sqrt41)+C, `
where C=C_1−C_2
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