Confused on algebraic step to derive a rule for partial fractions

Question

I am trying to understand the partial fraction identity $$\frac{5x+4}{(x^{2}+9)(x-1)} = \frac{A}{x-1} +\frac{Bx + C}{(x^{2}+9)}$$ I transformed it to be equal to $$\frac{A}{x-1} + \frac{B}{x+3i} + \frac{C}{x-3i} = \frac{A}{x-1} + \frac{Bx+C}{x^{2}+9}$$ I changed $$\frac{A}{x-1} + \frac{B}{x+3i} + \frac{C}{x-3i}$$ to be $$A(x+3i)(x-3i) + B(x-1)(x-3i)+C(x-1)(x+3i)$$ How do I proceed from this step to make it look like $$\frac{A}{x-1} +\frac{Bx + C}{x^{2}+9}?$$

Answer 1

$$\frac{A}{x-1} + \frac{B}{x+3i} + \frac{C}{x-3i} = \frac{ 5x+2}{(x^2+9) (x-1) }$$ $$ \implies A(x+3i)(x-3i) + B(x-1)(x-3i)+C(x-1)(x+3i)=5x+2$$ $$ \implies (A+B+C)x^2 + [ -B(1+3i) -C(1-3i) ] x+ +[ 9A+3i(B-C)] =5x+2$$

This is an identity in $x$ so we can equate coefficients of powers of $x$ we require: :: $$A+B+C=0$$ $$ -B(1+3i) -C(1-3i) =5 $$ $$9A+3i(B-C)=2$$

so $$B+C=-A$$ $$ B+C = -3i(B-C) -5 \implies 3i(B-C) = A-5 $$ $$9A+A-5=2 \implies A=\frac{7}{10}$$ $$ B-C = \frac{1}{3i} \left( -\frac{43}{10} \right) =i \left( \frac{43}{30} \right) $$ $$ B+C = -\frac{7}{10} $$

$$ B= -\frac{7}{20} + i \left( \frac{43}{60} \right) $$ $$ C= -\frac{7}{20} - i \left( \frac{43}{60} \right) $$

Putting it all together and using the expression given in my comment :

$$\frac{ 5x+2}{(x^2+9) (x-1) } =\frac{\frac{7}{10}}{x-1} + \frac{-\frac{7}{20} + i \left( \frac{43}{60} \right) }{x+3i} + \frac{-\frac{7}{20} - i \left( \frac{43}{60} \right) }{x-3i} \\= \frac{\frac{7}{10}}{x-1} + \frac{-\frac{7}{10} x + \left( \frac{43}{10} \right) }{x^2+9} $$