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प्रश्न
Find: `int(x+3)sqrt(3-4x-x^2dx)`
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उत्तर
`I=int(x+3)sqrt(3-4x-x^2)dx`
Let (x + 3) = `A*d/dx(3-4x-x^2)+B `
⇒ x + 3 = A(−2x − 4) + B
⇒ x + 3 = −2Ax − 4A + B
∴ −2A = 1
⇒ A = `-1/2`
⇒ −4A + B = 3
⇒ `-4(-1/2)+B=3`
⇒ B = 1
`:.I=int[-1/2*d/dx(3-4x-x^2)+1]sqrt(3-4x-x^2dx)`
`=1/2intd/dx(3-4x-x^2)sqrt(3-4x-x^2)dx+intsqrt(3-4x-x^2-4+4)dx`
`=1/2((3-4x-x^2)^(3/2)/(3/2))+intsqrt(7-(x+2)^2)dx`
`= -(3-4x-x^2)^(3/2)/3+(x+2)/2sqrt(7-(x+2)^2)+7/2sin^(-1)((x+2)/sqrt7)+C`
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