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.
 A: Split the fraction. 
$$\color{blue}{\dfrac{5 }{(x+1)(x^2 + 4)} = \dfrac{P(x)}{(x+1)}+\dfrac{Q(x)}{(x^2+4)}}$$
The largest partial fraction's denominator is an order 2 polynomial, so the order of the partial fractions' numerators must become order 2 when cross multiplied with the other partial fractions' denominators.  Ie: P(X) is order 0, Q(X) is order 1.
$$\color{blue}{\dfrac{5 }{(x+1)(x^2 + 4)} = \dfrac{A}{(x+1)}+\dfrac{Bx+C}{(x^2+4)}}$$
Now that you know what the target looks like, perform the crossmultiplication.
$$\color{blue}{5 = A(x^2+4)+(Bx+C)(x+1)}$$
Expand the polynomial so we can compare the coefficients.
$$\color{blue}{5 = (A+B)x^2 + (B+C)x + (4A+C)}$$
Which allows us to conclude
$$\color{blue}{A=-B, C=-B, 5= 4A+C \implies B=-1, A=1, C=1}$$
So we have our partial fraction expansion.
$$\color{blue}{\therefore \dfrac{5 }{(x+1)(x^2 + 4)} = \dfrac{1}{(x+1)}+\dfrac{(x+1)}{(x^2+4)}}$$
Next use: $\color{blue}{\dfrac{\mathrm{d}}{\mathrm{d}x}\ln(x+a) = \dfrac{1}{x+a}}$, $\color{blue}{\dfrac{\mathrm{d}}{\mathrm{d}x}\ln(x^2+b^2)= 2\dfrac{x}{x^2+b^2}}$ and $\color{blue}{\dfrac{\mathrm{d}}{\mathrm{d}x} \arctan \dfrac{x}{c} = \dfrac{c}{x^2+c^2}}$
A: 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 $x$.
$$
\frac{5}{(x + 1)(x^2 + 4)} = \frac{1}{x + 1} + \frac{1 - x}{x^2 + 4} = \frac{1}{x + 1} + \frac{1}{x^2 + 4} - \frac{x}{x^2 + 4}
$$
A: Note as in the comments you will have
$$
\dfrac{5 }{(x+1) (x^2 + 4)  }
= \frac{A}{x+1} + \frac{Bx + C}{x^2+4}
$$
Cross multiplying and solving the equations gives
$$
\dfrac{5 }{(x+1) (x^2 + 4)  }
= \frac{1}{x+1} - \frac{x - 1}{x^2+4}
$$
Which is easier to integrate. A sneakier way than solving the equations is the following
$$
\frac{5}{(x+1)(x^2+4)}
=
\frac{(x^2 + 4) + (1 - x^2)}{(x+1)(x^2+4)}
= \frac{1}{x+1} + \frac{(1-x)(1+x)}{(x+1)(x^2+4)}
$$
As wanted.
