91Ó°ÊÓ

The rational expression is,

$\frac{1}{(x+1)(x^2+4)}$.

Decompose the rational fraction, we get,

$\frac{1}{(x+1)(x^2+4)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+4}...........\left(1\right)$

Multiplying both sides by $(x+1)(x^2+4)$,

$$\begin{gathered} 1=A(x^2+4)+(Bx+C)(x+1) \\ 1=A{x}^2+4A+B{x}^2+Bx+Cx+C \\ 1=(A+B)x^2+(B+C)x+(4A+C)..............\left(2\right) \end{gathered}$$

Equating the coefficients with the like powers to get,

$$\begin{gathered} A+B=0.........\left(3\right) \\ B+C=0........\left(4\right) \\ 4A+C=1........\left(5\right) \end{gathered}$$

From equation $\left(3\right)$, we get,

$$\begin{gathered} A+B=0 \\ ⇒A=-B \end{gathered}$$

Inputting the value in $\left(5\right)$. we get,

$-4B+C=1.....\left(6\right)$

role="math" localid="1647416500158" $B+C=0......\left(4\right)$

Solving the equations we get,

$B=-\frac{1}{5}$

$C=\frac{1}{5}$

$A=\frac{1}{5}$

The partial fraction decomposition is,

role="math" localid="1647416720742" $\frac{1}{(x+1)(x^2+4)}=\frac{{\displaystyle \frac{1}{5}}}{(x+1)}+\frac{-{\displaystyle \frac{1}{5}}x+{\displaystyle \frac{1}{5}}}{x^2+4}$

$$\begin{gathered} =\frac{{\displaystyle \frac{1}{5}}}{x+1}+\frac{{\displaystyle \frac{1}{5}}(-x+1)}{x^2+4} \\ =\frac{1}{5(x+1)}+\frac{-x+1}{5(x^2+4)} \end{gathered}$$

Adding the rational expressions,

$\frac{1}{5(x+1)}+\frac{-x+1}{5(x^2+4)}=\frac{x^2+4+(x+1)(-x+1)}{5(x+1)(x^2+4)}$

$$\begin{gathered} =\frac{x^2+4-x^2-x+x+1}{5(x+1)(x^2+4)} \\ =\frac{5}{5(x+1)(x^2+4)} \\ =\frac{1}{(x+1)(x^2+4)} \end{gathered}$$

This is true.