(4x³ + 8x² + 6x + 1)/(4x² + 4x + 1)

= ((4x³ + 4x² + x) + (4x² + 4x + 1) + x)/(4x² + 4x + 1)

= ( x(4x² + 4x + 1) + 1(4x² + 4x + 1) + x)/(4x² + 4x + 1)

= x + 1 + x/(4x² + 4x + 1)

= x + 1 + x/( (2x)² + 2(1)(2x) + 1²)

= x + 1 + x/(2x + 1)²

(Or you can do long division, whichever you prefer)



= x + 1 + ½(2x)/(2x + 1)²

= x + 1 + ½( (2x + 1) - 1)/(2x + 1)²

= x + 1 + ½( 1/(2x + 1) - 1/(2x + 1)² )

= x + 1 + 1/(2(2x + 1)) - 1/(2(2x+1)²)


Or


let x/(2x + 1)² = A/(2x + 1) + B/(2x + 1)²


Multiply both sides by (2x + 1)²,

x = A(2x + 1) + B

Sub x = -½,

-½ = B


Sub x = 0, B = -½

0 = A -½

A = ½

So,

(4x³ + 8x² + 6x + 1)/(4x² + 4x + 1)

= x + 1 + 1/(2(2x + 1)) - 1/(2(2x+1)²)

ohh but how to find A and B？

Refer to the above edited working

First part is by observation and factoring out terms. Second part is the usual partial fractions method.

You can actually skip long division if you observe that the terms can be rearranged and factored out as above.

There are also non Long division approaches for the decomposition. A degree 3 polynomial divided by a degree 2 polynomial yields a degree 1 quotient and a fraction in which the numerator is at least one power less than the denominator (dividend).

You can let the fraction be equal to Ax + B + (Cx + D) / (4x2 + 4x + 1) and solve the question the identity way as well.

The resulting proper fraction is x / (2x + 1)2 which can also be transformed by observing that the numerator can be written as 0.5 (2x + 1) - 0.5.