ज्ञात करें `int (3x+5)/(x^3 -x^2 -x+1) dx`

Explanation

`(3x+5)/(x^3-x^2-x+1)=(3x+5)/(x^2(x-1)-(x-1))=(3x+5)/((x-1)(x^2-1))=(3x-5)/((x-1)^2(x+1))`
माना की `(3x+5)/((x-1)^2(x+1))=A/(x-1)+B/((x-1)^2)+C/(x+1)`
`rArr" "3x+5 -=A(x-1)(x+1) +B(x+1) +C(x-1)^2" ...(1)"`
(1) में `x = 1, -1` रखने पर,
`8=B(2)rArr B=4` तथा `2=C(4) rArrC=1/2`
अब, `(3x+5)/(x^3-x^2 -x+1)=-1/2""1/((x-1))+4/((x-1))+1/2""1/((x+1))`
`:." "int (3x+5)/(x^3 -x^2 -x+1)dx = -1/2 int (dx)/(x-1) +4 int (dx)/((x-1)^(2))+1/2int (dx)/(x+1)`
`=-1/2 log|x-1| - 4/(x-1)+1/2 |x+1|+c`