`(x^(2)+x+1)/((x+1)^(2)(x+2))`

Explanation

`int(x^(2)+x+1)/((x+1)^(2)(x+2))`
माना `(x^(2)+x+1)/((x+1)^(2)(x+2))=(A)/(x+1)+(B)/((x+1)^(2))+(C)/((x+2))" …(1)"`
`rArr x^(2)+x+1=A(x+1)(x+2)+B(x+2)+C(x+1)^(2)`
`{:(x-1,"तो ", 1-1+1=0+B(-1+2)+0 ,rArr., B=1),(x=-2,"तो ",4-2+1=0+0+C(1),rArr., C=3):}`
`x^(2)` के गुणकों को बराबर रखने पर ,
`1=A+C" "rArr" "A=-2`
`therefore" "(x^(2)+x+1)/((x+1)^(2)(x+2))=(-2)/(x+1)+(1)/((x+1)^(2))+(3)/((x+2))`
`therefore" "int(x^(2)+x+1)/((x+1)^(2)(x+2))dx`
`=-2 int(dx)/(x+1)+int(dx)/((x+1)^(2))+3int(d)/((x+2))`
`=-2log|x+1|-(1)/((x+1))+3log |x+2|+C`