<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] 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.
</s>