<s>[INST] Partial fraction integral Question:
$\int \dfrac{5 }{(x+1) (x^2 + 4)  } dx $ 
Thought process: 
I'm treating it as a partial fraction since it certainly looks like one.
I cannot seem to solve it besides looking at it in the "partial fraction" way.
My work:
1) Focus on the fraction part first ignoring the $\int $ and $dx$ for the moment.
Multiply $(x+1) (x^2+4)$ on both sides of the equation and get: 
$5$ =  $\dfrac{A }{(x+1) } $ +  $\dfrac{Bx+C }{(x^2+4) } $ 
Note: the x^2 + 4 is irreducible which explains the Bx+c as the numerator.
$ 5 = A(x^2+ 4) + Bx + C(x+1) $
I tried x = -1 which knocks out C:
$5 = A5 + - B $
I also tried x = 0 which knocks out B
$5 = A4 + C$ 
A is a lot harder to knock out since the squared changes the picked value to be positive.
I decided to add the two found equation together and get
$10 = A9 - B + C $
Now I am officially stuck at this step.
 [/INST] Your equation is incorrect--you have some algebra mistakes that are causing you to not get a solution (which is obvious unless you start plugging in values to $x$):
$$
\frac{5}{(x + 1)(x^2 + 4)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 4}
$$
Now cross multiply to get the numerator:
\begin{align}
5 =& A(x^2 + 4) + (Bx + C)(x + 1) = Ax^2 + 4A + Bx^2 + Bx + Cx + C \\
=& (A + B)x^2 + (B + C)x + (4A + C)
\end{align}
This yields exactly three equations for three unkowns.  You don't plug in values, you match coefficients: $5 = 0*x^2 + 0 * x + 5$ therefore you have:
\begin{align}
(A + B)x^2 =& 0*x^2& &\longrightarrow &A + B = 0 \\
(B + C)x =& 0*x&& \longrightarrow &B + C = 0  \\
4A + C =& 5 && \longrightarrow & 4A + C = 5 \\
\end{align}
Solving those equations yields:
\begin{align}
A + B = 0 &\longrightarrow A = -B \\
B + C = 0  & \longrightarrow C = -B = A\\
4A + C = 5 &\longrightarrow 4A + A = 5 \longrightarrow A = 1, B = -1, C = 1\\
\end{align}
Solve those equations and you will get the correct partial fraction decomposition--no need to plug in arbitrary values of $x$.
$$
\frac{5}{(x + 1)(x^2 + 4)} = \frac{1}{x + 1} + \frac{1 - x}{x^2 + 4} = \frac{1}{x + 1} + \frac{1}{x^2 + 4} - \frac{x}{x^2 + 4}
$$
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