1.

`int(x+3)sqrt(3-4x-x^(2))dx` का मान ज्ञात कीजिए ।

Answer» `int(x+3)sqrt(3-4x-x^(2))dx`
अब माना कि `x(x+3)=A.(d)/(dx)(3-4x-x^(2))+B`
`rArr" "(x+3)=A(-4-2x)+B" …(1)"`
अब समीकरण (1 ) के दोनों पक्षों में x की घातों की तुलना करने पर
`-2A=1 rArr A=-(1)/(2)`
तथा `-4A+B=3 rArr B=1`
अब `int(x+3)sqrt(3-4x-x^(2))dx`
`=int{-(1)/(2)(-4-2x)+1}.sqrt(3-4x-x^(2))dx`
`=-(1)/(2)int(-4-2x)sqrt(3-4x-x^(2))dx+intsqrt(3-4x-x^(2))dx`
यदि `3-4x-x^(2)=t" व "dt=(-4-2x)dx`
`=-(1)/(2)intsqrtt dt +int sqrt(7-(4x+x^(2)+4))dx`
`=(-(1)/(2)xxt^(3//2)xx(2)/(3))+intsqrt((sqrt7)^(2)-(x+2)^(2))dx`
`=-(1)/(3)(3-4x-x^(2))^(3//2)+(1)/(2)(x+2)sqrt(3-4x-x^(2))+(7)/(2)sin^(-1)((x+2)/(sqrt7))+c`
`[" यहाँ "intsqrt(a^(2)-x^(2))dx=(x)/(2)sqrt(A^(2)-x^(2))+(a^(2))/(2)sin^(-1).(x)/(a)+c]`


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