I'm self-studying from Stroud & Booth's amazing "Engineering Mathematics", but am stuck on a problem from the "Partial Fractions" chapter.
I've been running around in circles trying to solve it, and just can't seem to get it right (although I understand the general mechanics around the solution process).
The expression that I need to split into partial fractions is:
$$\frac{9x^2+48x+18}{(2x+1)(x^2+8x+3)}$$
Can anyone please help me out?
(Assuming you want to split it into partial fractions)
Let $$\frac{9x^2+48x+18}{(2x+1)(x^2+8x+3)} = \frac{a}{2x+1} + \frac{bx+c}{x^2+8x+3}$$
By inspection,
At $x=-\frac{1}{2}$,
$$\frac{9/4-48/2+18}{1/4-8/2+3} = a\implies a = 5$$
At $x=0$
$$\frac{0+0+18}{(0+1)(0+0+3)} = \frac{5}{0+1}+\frac{c}{0+0+3}\implies c =3 $$
At $x=1$
$$\frac{9+48+18}{3(1+8+3)} = \frac{5}{3}+\frac{b+3}{1+8+3}\implies b=2$$
So, $$\frac{9x^2+48x+18}{(2x+1)(x^2+8x+3)} = \frac{5}{2x+1} + \frac{2x+3}{x^2+8x+3}$$