(3x^2+7) / ( (x-1)(x^2+2x+5) )
Your set up will look like this,
$\displaystyle \frac{3x^2+7}{(x-1)(x^2+2x+7}=\frac{A}{(x-1)}+\frac{Bx+C}{(x^2+2x+5)}$
Clearing the denominators gives us,
$\displaystyle 3x^2+7=A(x^2+2x+5)+(Bx+c)(x-1)$
Test a couple of values of x like x=1 to solve for A , B and C
Your solution should come out as,
$\displaystyle A=\frac{5}{4} \\\ B= \frac{7}{4} \\\ and \\\ C=-\frac{3}{4}$
Which would give you,
$\displaystyle \frac{5}{4(x-1)}+\frac{7x+3}{4(x^2+2x+5)}$
Hope that helps!!