Precalculus, Chapter 7, 7.4, Section 7.4, Problem 49

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 geta=-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).

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