`(x^4 + 2x^3 + 4x^2 + 8x + 2)/(x^3 + 2x^2 + x)` Write the partial fraction decomposition of the improper rational expression.

A bit more extension for my above solution 

Now in the above expression we need to simplify the

`(2x-1)/(x^3 + 2x^2 + x)`
It is as follows

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

`(2x-1)/(x(x+1)^2)= (a/x) + (b/(x+1))+(c/(x+1)^2))`

on simplification we get
`(2x-1)= (a(x+1)^2)+(bx(x+1))+cx`

As the roots of the denominator `(x(x+1)^2)` are` 0 , -1` . We can solve the unknown parameters by
plugging the values of `x` .

when `x=0` , we get
`a=-1 `
when `x=(-1)` we get
`c=3`

As we know the `a,c` values , we can find the value of `b` as

`2x-1 = (-1)(x+1)^2 + bx(x+1)+3x`
`2x-1 = bx^2+x+bx-x^2-1`
`2x-1 = x^2(b-1)+x(b+1)-1`
on comparing we get
`b+1 =2`
=> `b=1 `
so, `(2x-1)/(x^3 + 2x^2 + x) = ((-1)/x)+(1/(x+1))+(3/(x+1^2))`
so, the partial fraction for

`(x^4 + 2x^3 + 4x^2 + 8x + 2)/(x^3 + 2x^2 + x) = x+ (3/x)+ (2x-1)/(x^3 + 2x^2 + x)`

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