` int[3x+5]/[x^3-x^2-x+1].dx`

Explanation

` int (3x+5)/(x^(3)-x^(2)-x+1)dx`
`=int(3x+5)/((x^(2)-1)(x-1))dx`
`=int(3x+5)/((x-1)(x+1)(x-1))dx`
`=int(3x+5)/((x-1)^(2)(x+1))dx`
`" Let " (3x+5)/((x-1)^(2)(x+1))=(A)/(x-1)+(B)/((x-1)^(2))+(C)/(x+1)`
` rArr (3x+5)/((x-1)^(2)(x+1))`
`A(x-1)(x+1)`
` =(+B(x+1)+C(x-1)^(2))/((x-1)^(2)(x+1))`
`rArr 3x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^(2)`
`x=1 " then " 3+5 =0 +B(2)+0 rArr B=4`
`x=-1 " then " -3 +5=0+0 +C(-2)^(2) rArr C=(1)/(2)`
`rArr ` Equating coefficients of `x^(2)`
0= A+C
`rArr A=-C =-(1)/(2)`
`:. int (3x+5)/((x-1)^(2)(x+1))dx`
`=int ((-(1)/(2))/(x-1)+(4)/((x-1)^(2))+((1)/(2))/(x+1))dx`
`=-(1)/(2) int(1)/(x-1) dx+4 int(1)/((x-1)^(2))dx`
`+(1)/(2) int(1)/(x+1)dx`
`=-(1)/(2) log |x-1| +4.((x-1)^(-2+1))/((-2+1))`
` +(1)/(2) Log|x+1|+C`
`=(1)/(2) log |(x+1)/(x-1)| -(4)/(x-1)+C`