ज्ञात करें `int ( 2x-5) sqrt( x^2 -4x+3) dx`

Explanation

माना कि `I=int (2x-5) sqrt(x^2 -4x+3)dx`
`=int (2x-4-1) sqrt(x^2 -4x+3)dx`
`=int ( 2x-4) sqrt(x^2 -4x +3)dx -int sqrt(x^2 -4x +3) dx`
`=I_(1)-I_(2)+c`
जहां `I_(1)=int sqrt(x^2 -4x+3) (2x-4)dx ` तथा `I_(2) int sqrt(x^2 -4x +3) dx`
`=I_(1)-I_(2)+c" ...(1)"`
जहां `I_(1) = int sqrt(x^2 -4x+3)(2x-4)dx` तथा `I_(2) = int sqrt(x^2 - 4x +3) dx`
`I_(1)` निकालना : `x^2 -4x+3 =t ` रखे, तो `(2x-4) dx = dt`
` :." "I_(1) = int sqrt(t) dt = t^(3//2)/(3//2) = 2/3(x^2-4x+3)^(3//2)" ..(2)"`
पुनः `I_(2) = int sqrt(x^2 -4x+3)dx = intsqrt((x-1)^2-1^2)dx`
`=((x-2)sqrt((x-2)^2-1^2))/2-1^2/2log | (x-2)+sqrt((x-1)^2 -1^2)|`
`=((x-2)sqrt(x^2-4x+3))/2-1/2"log "|(x-2)+sqrt(x^2 -4x+3)|" ...(3)"`
`:." "` (1), (2) तथा (3) से, हमे मिलता है
`I=2/3(x^2 -4x+3)^(3//2)-1/2(x-2)sqrt(x^2-4x+3)+1/2"log"|x-2+sqrt(x^2-4x+3)|+c`