Use partial fractions to find the integral. Find the integral using partial factions.
$$\int\frac{(2x^2+5x+3)}{(x-1)(x^2+4)}\,dx$$
So do I do...$$\frac{2x^2+5x+3}{(x-1)(x^2+4)}=\frac{A}{x-1} + \frac{Bx+C}{x^2+4},$$
then get 
\begin{align*}
2x^2+5x+3 &= A(x^2+4)+(Bx+C)(x-1) \\
2x^2+5x+3 &= Ax^2+4A+Bx^2-Bx+Cx-C?
\end{align*}

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 $$2(1)^2 + 3(1) + 5 = A((1)^2 + 4) + 0$$
$$2 + 3 + 5 = 5A$$
$$A=2$$
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 $B = 0, C = 5$.
