# Decomposing to Partial Fractions

I have this new scenario to decompose:

$$\dfrac{4x^2+2x}{(x^2+1)^2}$$

From my assessment, I that I don't have to use the long division.

$(i).... \dfrac{4x^2+2x}{(x^2+1)^2} = \dfrac{4x^2+2x}{(x^2+1)(x^2+1)}=\dfrac{Ax+B}{(x^2+1)}+ \dfrac{Cx+D}{(x^2+1)^2}$

When I multiply with LCD I get:

$(ii)....4x^2+2x = (Ax+B)(x^2+1)+Cx+D$

$(iii)....==>4x^2+2x = Ax^3+Ax+Bx^2+B+Cx+D$

(iv)....collect like terms:$4x^2+2x =Ax^3+Bx^2+Ax+Cx+B+D$

(v)....$0x^3+4x^2+2x+0 =(Ax^3)+(Bx^2)+(A+C)x+D$

(vi)...equate coefficients to get $A=0, B=4, D=0, C=2$

so the solution be:

$\dfrac{4x^2+2x}{(x^2+1)^2} = \dfrac{0x+4}{(x^2+1)}+ \dfrac{2x+0}{(x^2+1)^2}$

decomposes to:

$\dfrac{4x^2+2x}{(x^2+1)^2} = \dfrac{4}{(x^2+1)}+ \dfrac{2x}{(x^2+1)^2}$

Please help me check this. I would be happy to get it right this time!

• "Help you to check that"?? Do the sum of fractions in the right side and you check whether it equals the left side! – DonAntonio Oct 3 '13 at 12:13
• Closing is an overreaction here. OP did almost everything themselves. – Lord_Farin Oct 3 '13 at 12:46
• I agree with Lord : to close a question like this one is way to harsh and silly (no offence). – DonAntonio Oct 3 '13 at 12:52

You erred in determining the value of D when you equated coefficients: Specifically you forgot to include $B+D = 0$, not just $D = 0$

(iv) $\quad 4x^2+2x =Ax^3+Bx^2+Ax+Cx+\color{red}{\bf B+D}$

(v) $\quad 0x^3+4x^2+2x+0 =(Ax^3)+(Bx^2)+(A+C)x+\color{red}{\bf (B+D)}$

So we have $B+ D = 0$, and hence, since $B= 4$, then $D= - 4$.

So the fraction decomposes like this:

$$\dfrac{4x^2+2x}{(x^2+1)^2} = \dfrac{4}{(x^2+1)}+ \dfrac{2x - 4}{(x^2+1)^2}$$

Note: The process you used in decomposing the rational function into partial fractions was "spot on", so you deserve credit for understanding the "rationale" for and logic of parial fraction decomposition. The only error here was a mismatch of the last coefficient, after having "dropped a variable" when moving from step (iv) to step (v).

• I see what you mean. I erred on line (v). I did not bring down the full variables. Many thanks. – Sylvester Oct 3 '13 at 12:52
• You're welcome, Sylvester! – Namaste Oct 3 '13 at 12:53
• @amWhy: Needs another TU! +1 – Amzoti Oct 3 '13 at 23:57