Answer
$\dfrac{\frac{1}{5}}{x+1}+\dfrac{\frac{-x}{5}+\frac{1}{5}}{x^2+4}$
Work Step by Step
We can notice from the given rational expression $\dfrac{A(x)}{B(x)}$ that $B(x)$ has only non-repeated irreducible quadratic factor.
So, we will write the given rational expression into partial fraction form as:
$$\displaystyle \frac{1}{(x+1)(x^2+4)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+4} \quad \quad(1)$$
Now, write the RHS with a common denominator.
\begin{align*}\frac{1}{(x+1)(x^2+4)}&=\frac{A(x^2+4)+(x+1)(Bx+C)}{(x+1)(x^2+4)}\\
\\\frac{1}{(x+1)(x^2+4)}&=\frac{Ax^2+4A+Bx^2
+Cx+Bx+C}{(x+1)(x^2+4)}
\\\frac{1}{(x+1)(x^2+4)}&=\frac{Ax^2+Bx^2+Bx+Cx+4A+C}{(x+1)(x^2+4)}
\end{align*}
Next, equate the numerators to obtain:
$$(A+B)x^2+(B+C)x+(4A+C)=1$$
Equate the coefficients of the polynomials on the LHS and RHS to obtain:
$$A+B=0 \quad \quad\ (2)
\\ B+C=0 \quad \quad (3)
\\4A+C=1\quad \quad(4)$$
Equation $(2)$ yields: $A=-B$.
Plug $A=-B$ into the Equation $(3)$ and then subtract equations $(3)$ from $(4)$ to solve for $A$. Thus, we have:
\begin{align*}(4A+C)-(-A+C)&=1-0\\
4A+C+A-C&=1\\
5A&=1\\
A&=\frac{1}{5}
\end{align*}
Hence, $B=-A=-\dfrac{1}{5}$
Finally, plug $B=-\frac{1}{5}$ into Equation $(3)$ obtain:
$$-\dfrac{1}{5}+C =0\implies C=\dfrac{1}{5}$$
Therefore, equation $(1)$ becomes:
$$\dfrac{1}{(x+1)(x^2+4)}=\dfrac{\frac{1}{5}}{x+1}+\dfrac{\frac{-x}{5}+\frac{1}{5}}{x^2+4}$$
