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प्रश्न
Find : `int(x+3)sqrt(3-4x-x^2dx)`
उत्तर
`I=int(x+3)sqrt(3-4x-x^2)dx`
Let (x+3)= `Ad/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[-1/2d/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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