Evaluate: $ \displaystyle\int \frac{9x^2 - 30x - 73}{(x - 4)(x^2 - x - 12)} \; dx$ Partial fractions

Question

Evaluate: $\; \displaystyle\int \frac{9x^2 - 30x - 73}{(x - 4)(x^2 - x - 12)} \; dx$

I couldn't find A, how to find A in the partial fractions below?

$$\int \frac{9x^2 - 30x - 73}{(x - 4)(x+3)(x-4)} \; dx $$

$$\int \frac{9x^2 - 30x - 73}{(x - 4)^2(x+3)} \; dx $$

Solving what's inside integral:

$$\frac{9x^2 - 30x - 73}{(x - 4)^2(x+3)} \; dx = \frac{A}{(x-4)} + \frac{B}{(x-4)^2} + \frac{C}{(x+3)} $$

->

$$ 9x^2 - 30x - 73 = A(x-4)(x+3) + B (x+3) + C(x-4)^2$$

$$x = 4, \;\; 9 \cdot 4^2 - 30 \cdot 4 - 73 = A \cdot 0 + B (4 + 3) + C \cdot 0 $$

$$B = \frac{-49}{7} = -7$$

$$x = -3, \;\; 9 \cdot (-3)^2 - 30 \cdot (-3) - 73 = A \cdot 0 + B \cdot 0 + C (-7)^2 $$

$$C = \frac{98}{49} = 2$$

I tried to find A by plugging in the now-known values of B and C, but to no avail. How do I find A?

Answer 1

Look at your second displayed equation, the one that starts with $9x^2-30x-73=$. The coefficients of $x^2$ on the left-hand side and right-hand side must match.

The coefficient of $x^2$ on the left-hand side is $9$. Imagine expanding the right-hand side, you will get $(A+C)x^2+\text{stuff}$.

So $9=A+C$. We know $C$, so we know $A$.

Remark: In principle, we are trying to find numbers $A$, $B$, and $C$ such that the polynomial $9x^2-30x-73$ is identically equal to the polynomial $A(x-4)(x+3)+B(x+3)+C(x-4)^2$.

One way to find such $A,B,C$ is to match coefficients. That gives us three linear equations in the three unknowns $A,B,C$, which we solve in the usual way. That can be tedious, and we can $B$ and $C$ in this case by using substitution tricks. We can also find $A$ by a substitution trick, for example by letting $x=0$. However, to find $A$ we instead went partly back to basics and used matching of coefficients.

Answer 2

Since $B=-7$ and $C=2$, we can say that $9x^2 -30x-73=A(x-4)(x+3)-7(x+3)+2(x-4)^2$. Try plugging in $0$ or any other value other than $4$ or $-3$ for $x$ and solving for $A$.