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*}
 A: 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 $B = 0, C = 5$.
A: Hint: Write out the fraction given as $\frac{A}{x-1}+\frac{Bx+C}{x^2+4}$ and equate the numerator to $2x^2+5x+3$. And the ensuing integral should be easy. 
A: Note that we can simply add zero: $2x^2 = 2(x^2 + 4) - 8 = 2x^2 + 8 - 8$. Then
$$\frac{2x^2 + 5x + 3}{(x - 1)(x^2 + 4)} = \frac{2(x^2 + 4) - 8 + 5x + 3}{(x - 1)(x^2 + 4)} = \frac{2}{x - 1} + \frac{5x - 5}{(x - 1)(x^2 + 4)}.$$
Do you think you can integrate
$$\frac{2x^2 + 5x + 3}{(x - 1)(x^2 + 4)}$$
now?
Hint:
$$\frac{5x - 5}{(x - 1)(x^2 + 4)} = \frac{5(x - 1)}{(x - 1)(x^2 + 4)} = \frac{5}{x^2 + 4}$$
and
$$\int \frac{1}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C, \quad a \ne 0.$$
