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

Explanation

`int(1)/(x-x^(3))dx=int(1)/(x(1-x^(2)))dx`
`=int(1)/(x(1-x)(1+x))dx" …(1)"`
माना `(1)/(x(1-x)(1+x))=(A)/(x)+(B)/((1-x))+(C)/((1+c))`
`rArr" "1=A(1+x)(1-x)+B(x)(1+x)+C(x)(1-x)`
`" "x=-1" तो "1=0+0+C(-1)(2)rArr" "C=-(1)/(2)`
`" "x=-1"तो "1=0+B(1)(2)+0rArr" "B=(1)/(2)`
`" "x=1"तो "1=A(1)(1)+0+0rArr" "A=1`
`therefore int(1)/(x(1-x^(2)))dx`
`=int(1)/(x)dx+(1)/(2)int(1)/(1-x)dx-(1)/(2)int(1)/(1+x)dx`
`=log|x|-(1)/(2)log||1-x|-(1)/(2)log|x+x|+C`
`=(1)/(2)log|(x^(2))/(1-x^(2))|+C`