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

Explanation

यदि `" "I=int(x+3)srt(3-4x-x^(2))dx`
अब `" "(d)/(dx)(3-4x-x^(2))=-4 - 2x`
अतः `" "x+3 =-(1)/(2)(-4 -2x)+1`
`therefore I=intsqrt(3-4x-x^(2)).{-(1)/(2)(-4-2x)+1}dx`
`=-(1)/(2)intsqrt(3-4x-x^(2))(-4 - 2x)dx+intsqrt(3-4x-x^(2))dx`
`=-(1)/(2)I_(1)+I_(2)`
यहाँ `I_(1)=intsqrt(3-4x-x^(2)).(-4-2x)dx`
`3-4x-x^(2)=t rArr -4-2xdx=dt`
`therefore" "I_(1)=intt^(1//2)dt=(t^(3//2))/(3//2)=(2)/(3)(3-4x-x^(2))^(3//2)" ...(2)"`
तथा `" "I_(2)=intsqrt(3-4x-x^(2))dx`
`=intsqrt(7-(x+2)^(2))dx`
`=intsqrt((sqrt7)^(2)-(x+2)^(2))dx`
`=((x+2)sqrt(3-4x-x^(2)))/(2)+(7)/(2)sin^(-1)((x+2)/(sqrt7))+c" ...(3)"`
अब, समीकरण (1 ), (2 ) व (3 ) से
`I=-(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`