It seems that you are stuck at the actual partial fraction decomposition, rather than at the integration. So let's pick up at $$2x^2+5x+3= A(x^2+4) + (Bx+C)(x-1)$$
While this must hold true for all values of $x$, certain values of $x$ will lead us to the values of $A,B,C$ more quickly...
$x = 1$ is a natural choice, as that will make the $(Bx+C)$ go away. We then have $$2(1)^2 + 3(1) + 5 = A((1)^2 + 4) + 0$$
$$2 + 3 + 5 = 5A$$
Then you could use complex x-values ($\pm 2i$ for example) and equate real and imaginary parts, but you could just pick a few other small (real) x-values for ease of computation.
So $x=2$ gives us $21 = 16 + 2B + C$, or $$2B +C = 5$$
Then $x = 3$ gives us $36 = 26 + (3B+C)(2)$, or $$3B+C = 5$$
Solving these last two equations simultaneously gives $B = 0, C = 5$.