Heaviside cover-up method

Factorize the expression in the denominator. Set up a partial fraction for each factor in the denominator. Apply the cover-up rule to solve for the new numerator of each partial fraction.

${\displaystyle {\frac {3x^{2}+12x+11}{(x+1)(x+2)(x+3)}}={\frac {A}{x+1}}+{\frac {B}{x+2}}+{\frac {C}{x+3}}}$

Set up a partial fraction for each factor in the denominator. With this framework we apply the cover-up rule to solve for A, B, and C.

1. D₁ is x + 1; set it equal to zero. This gives the residue for A when x = −1.

2. Next, substitute this value of x into the fractional expression, but without D₁.

3. Put this value down as the value of A.

Proceed similarly for B and C.

D₂ is x + 2; For the residue B use x = −2.

D₃ is x + 3; For residue C use x = −3.

Thus, to solve for A, use x = −1 in the expression but without D₁:

${\displaystyle {\frac {3x^{2}+12x+11}{(x+2)(x+3)}}={\frac {3-12+11}{(1)(2)}}={\frac {2}{2}}=1=A.}$

Thus, to solve for B, use x = −2 in the expression but without D₂:

${\displaystyle {\frac {3x^{2}+12x+11}{(x+1)(x+3)}}={\frac {12-24+11}{(-1)(1)}}={\frac {-1}{(-1)}}=+1=B.}$

Thus, to solve for C, use x = −3 in the expression but without D₃:

${\displaystyle {\frac {3x^{2}+12x+11}{(x+1)(x+2)}}={\frac {27-36+11}{(-2)(-1)}}={\frac {2}{(+2)}}=+1=C.}$

Thus,

${\displaystyle {\frac {3x^{2}+12x+11}{(x+1)(x+2)(x+3)}}={\frac {1}{x+1}}+{\frac {1}{x+2}}+{\frac {1}{x+3}}}$

When factors of the denominator include powers of one expression we

1. Set up a partial fraction for each unique factor and each lower power of D;
2. Set up an equation showing the relation of the numerators if all were converted to the LCD.

From the equation of numerators we solve for each numerator, A, B, C, D, and so on. This equation of the numerators is an absolute identity, true for all values of x. So, we may select any value of x and solve for the numerator.

${\displaystyle {\frac {3x+5}{(1-2x)^{2}}}={\frac {A}{(1-2x)^{2}}}+{\frac {B}{1-2x}}}$

Here, we set up a partial fraction for each descending power of the denominator. Then we solve for the numerators, A and B. As ${\displaystyle (1-2x)}$ is a repeated factor, we now need to find two numbers, as so we need an additional relation in order to solve for both. To write the relation of numerators the second fraction needs another factor of ${\displaystyle (1-2x)}$ to convert it to the LCD, giving us ${\displaystyle 3x+5=A+B(1-2x)}$. In general, if a binomial factor is raised to the power of ${\displaystyle n}$, then ${\displaystyle n}$ constants ${\displaystyle A_{k}}$ will be needed, each appearing divided by successive powers, ${\displaystyle (1-2x)^{k}}$, where ${\displaystyle k}$ runs from 1 to ${\displaystyle n}$. The cover-up rule can be used to find ${\displaystyle A_{n}}$, but is still ${\displaystyle A_{1}}$ that is called the residue. Here, ${\displaystyle n=2}$, ${\displaystyle A=A_{2}}$, and ${\displaystyle B=A_{1}}$

To solve for ${\displaystyle A}$ :

${\displaystyle A}$ can be solved by setting the denominator of the first fraction to zero, ${\displaystyle 1-2x=0}$.

Solving for ${\displaystyle x}$ gives the cover-up value for ${\displaystyle A}$: when ${\displaystyle x=1/2}$.

When we substitute this value, ${\displaystyle x=1/2}$, we get:

${\displaystyle 3\left({\frac {1}{2}}\right)+5=A+B(0)}$

${\displaystyle A={\frac {3}{2}}+5={\frac {13}{2}}}$

To solve for ${\displaystyle B}$ :

Since the equation of the numerators, here, ${\displaystyle 3x+5=A+B(1-2x)}$, is true for all values of ${\displaystyle x}$, pick a value for ${\displaystyle x}$ and use it to solve for ${\displaystyle B}$.

As we have solved for the value of ${\displaystyle A}$ above, ${\displaystyle A=13/2}$, we may use that value to solve for ${\displaystyle B}$.

We may pick ${\displaystyle x=0}$ , use ${\displaystyle A=13/2}$ , and then solve for ${\displaystyle B}$ :

${\displaystyle {\begin{aligned}3x+5&=A+B(1-2x)\\0+5&={\frac {13}{2}}+B(1+0)\\{\frac {10}{2}}&={\frac {13}{2}}+B\\-{\frac {3}{2}}&=B\\\end{aligned}}}$

We may pick ${\displaystyle x=1}$ , Then solve for ${\displaystyle B}$ :

${\displaystyle {\begin{aligned}3x+5&=A+B(1-2x)\\3+5&={\frac {13}{2}}+B(1-2)\\8&={\frac {13}{2}}+B(-1)\\{\frac {16}{2}}&={\frac {13}{2}}-B\\B&=-{\frac {3}{2}}\end{aligned}}}$

We may pick ${\displaystyle x=-1}$ . Solve for ${\displaystyle B}$ :

${\displaystyle {\begin{aligned}3x+5&=A+B(1-2x)\\-3+5&={\frac {13}{2}}+B(1+2)\\{\frac {4}{2}}&={\frac {13}{2}}+3B\\-{\frac {9}{2}}&=3B\\-{\frac {3}{2}}&=B\end{aligned}}}$

Hence,

${\displaystyle {\frac {3x+5}{(1-2x)^{2}}}={\frac {13/2}{(1-2x)^{2}}}+{\frac {-3/2}{(1-2x)}},}$

or

${\displaystyle {\frac {3x+5}{(1-2x)^{2}}}={\frac {13}{2(1-2x)^{2}}}-{\frac {3}{2(1-2x)}}}$