फलन `int(2x-5)sqrt((2+3x-x^(2)))dx` का x के सापेक्ष समाकलन कीजिए ।

Explanation

माना `" "I=int(2x-5)sqrt((2+3x-x^(2)))dx`
माना `" "(2x-5)=A(d)/(dx)(2+3x-x^(2))+B`
`=A(3-2x)+B`
`2x-5=-2Ax+(3A+B)`
दोनों पक्षों में समान घातीय पदों के गुणांकों की तुलना करने पर
`-2A=2`
`therefore" "A=-1`
तथा `" "3A+B=-5`
`rArr" "B=-3A-5`
`therefore" "B=-2`
`therefore I=int[-(3x-2x)sqrt((2x+3x-x^(2)))-2sqrt(2+3x-x^(2))]dx`
`=-int(3-2x)sqrt((2+3x-x^(2)))dx-2intsqrt((2+3x-x^(2)))dx`
`=-int(3-2x)sqrt((2+3x-x^(2)))dx-2intsqrt(2-(x^(2)-3x+(9)/(4)-(9)/(4)))dx`
`" "` (पूर्ण वर्ग बनाने पर )
`=-((2+3x-x^(2))^((1)/(2)+1))/((1)/(2)+1)-2intsqrt((17)/(4)-(x-(3)/(2))^(2))dx`
`=-(2)/(3)(2+3x-x^(2))^((3)/(2))-2intsqrt(((sqrt(17))/(2))^(2)-(x-(3)/(2))^(2))dx`
माना `(sqrt(17))/(2)=a` तथा `x-(3)/(2)=t therefore dx=dt`
`therefore" "I=-(2)/(3)(2+3x-x^(2))^(3//2)-2intsqrt((a^(2)-t^(2)))dt`
`=-(2)/(3)(2+3x-x^(2))^(3//2)-2[(t)/(2)sqrt(a^(2)-t^(2))+(a^(2))/(2)sin^(-1)((t)/(a))]`
`=-(2)/(3)(2+3x-x^(2))^(3//2)+(3-2x)/(2)sqrt(2+3x-x^(2))-(17)/(4)sin^(-1)(2x-3)/(sqrt(17))`