Express 14x²+13x/(4x²+4x+1)(x-1) as partial fractions.

Question

Express

[latex]\frac{14 x^2+13 x}{\left(4 x^2+4 x+1\right)(x-1)}[/latex]

as partial fractions.

Accepted Answer

The denominator is factorized to (2x + 1)² (x — 1). The repeated factor, (2x + 1)² , produces partial fractions of the form

[latex]\frac{A}{2 x+1}+\frac{B}{(2 x+1)^2}[/latex]

The factor, (x — 1), produces a partial fraction of the form [latex]\frac{C}{x-1}[/latex]. So

[latex]\frac{14 x^2+13 x}{\left(4 x^2+4 x+1\right)(x-1)}=\frac{14 x^2+13 x}{(2 x+1)^2(x-1)}=\frac{A}{2 x+1}+\frac{B}{(2 x+1)^2}+\frac{C}{x-1}[/latex]

Multiplying both sides by (2x + 1)² (x — 1) gives

[latex]14 x^2+13 x=A(2 x+1)(x-1)+B(x-1)+C(2 x+1)^2[/latex] (1.10)

The unknown constants A, B and C can now be found.
Evaluating Equation (1.10) with a = 1 gives

[latex]27=C(3)^2[/latex]

from which

C = 3

Evaluating Equation (1.10) with x = —0.5 gives

[latex]14(-0.5)^2+13(-0.5)=B(-0.5-1)[/latex]

from which

B = 2

Finally, comparing the coefficients of x² on both sides of Equation (1.10) we have

14 = 2A +4C

Since we already have C = 3 then

A = 1

Hence we see that

[latex]\frac{14 x^2+13 x}{\left(4 x^2+4 x+1\right)(x-1)}=\frac{1}{2 x+1}+\frac{2}{(2 x+1)^2}+\frac{3}{x-1}[/latex]

Question 1.40: Express 14x²+13x/(4x²+4x+1)(x-1) as partial fractions....