`(x^(3)+x+1)/(x^(2)-1)`

Explanation

`(x^(3)+x+1)/(x^(2)-1)=x+(2x+1)/(x^(2)-1)`
`therefore" "int(x^(3)+x+1)/(x^(2)-1)dx=intx dx +int(2x+1)/(x^(2)-1)dx`
`" "=intxdx+int(2x+1)/((x+1)(x-1))dx`
माना `(2x+1)/((x+1)(x-2))=(A)/((x+1))+(B)/((x-1))`
`rArr (2x+1)/((x+1)(x-1))=(A(x-1)+B(x+1))/((x+1)(x-1))`
`rArr 2x+1=A(x-1)+B(x+1)`
`{:(x=1,"तो ",2+1=0+B(1+1),rArr.,B=(3)/(2)),(x=-1,"तो ",-2+1=A(-1-1)+0,rArr.,A=(1)/(2)):}`
`therefore" "int(x^(2)+x+1)/(x^(2)-1)dx=intxdx+int(A)/((x+1))dx+int(B)/((x-1))dx`
`=intxdx+(1)/(2)int(1)/((x+1))dx+(3)/(2)int(1)/((x-1))dx`
`=(x^(2))/(2)+(1)/(2)log|x+1|+(3)/(2)log|x-1|+C`