ज्ञात करें `int (2x-1)/((x-1)(x+2)(x-3)) dx `

Explanation

माना की `I=int (2x-1)/((x-1)(x+2)(x-3)) dx `
माना की `(2x-1)/((x-1)(x+2)(x-3))-=A/(x-1)+B/(x+2)+C/(x-3)`
`:." "2x-1 = A(x+2)(x-3)+B(x-1)(x-3)+C (x-1)(x+2)" ...(1)"`
(1) में `x =1, -2` तथा 3 रखने पर हमे मिलता है,
`1=A(3)(-2)rArr A = -1/6 ," "-5 =B (-3)(-5) rArr B = -1/3`
`5=C(2)(5) rArr C=1/2`
`:." "int (2x-1)/((x-1)(x+2)(x-3))dx = int [-1/6""1/((x-1))""-1/3""1/((x+@)) +1/2""1/((x-3))]dx`
अतः `I = 1/6 log | x-1| -1/3 log |x +2| 1/2 log | x-3 |+c`