(3x^2+7) / ((x-1)(x^2+2x+5))
I'm going to verify your answer:
$\displaystyle \frac{3x^2+7}{(x-1)(x^2+2x+5)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+2x+5}$
Multiply both sides by the denominator of the left term:
$\displaystyle 3x^2+7=A(x^2+2x+5)+(Bx+C)(x-1)$
Now, we get:
$\displaystyle 3x^2+7=Ax^2+2Ax+5A+Bx^2-Bx+Cx-C$
$\displaystyle 3x^2+7=(A+B)x^2+(2A-B+C)x+5A-C$
We now compare the coefficients:
$\displaystyle \begin{aligned}
A+B &=3 \\
2A-B+C &=0 \\
5A-C &=7
\end{aligned}
$
From the first equation, we see that $\displaystyle B=3-A$. From the third equation, we see that $\displaystyle C=5A-7$. Now substitute these values for B and C into the second equation:
$\displaystyle 2A-(3-A)+(5A-7)=0 \implies 8A-10=0\implies \color{red}\boxed{A=\frac{5}{4}}$
Thus, $\displaystyle \color{red}\boxed{C=-\frac{3}{4}}$ and $\displaystyle \color{red}\boxed{B=\frac{7}{4}}$
Thus the PFD is:
$\displaystyle \frac{\frac{5}{4}}{x-1}+\frac{-\frac{3}{4}x+\frac{7}{4}}{x^2+2x+5}\implies \color{red}\boxed{\frac{5}{4(x-1)}+\frac{7-3x}{4(x^2+2x+5)}}$.
It's Good!!