To calculate: The partial decomposition of x2−2x−21x3+7x

Calculation:
Consider the provided expression:
${\displaystyle \frac{x^2-2x-21}{x^3+7x}}$
Here, ${\displaystyle f\left(x\right)=x?-2x-21\text{and}g\left(x\right)=x^{\circ }+Tx}$
Factorize ${\displaystyle g\left(x\right):}$
${\displaystyle g\left(x\right)=x^3+7x}$
${\displaystyle =x(x^2+7)}$
So, the expression can be written as:
${\displaystyle \frac{x^2-2x-21}{x^3+7x}=\frac{x^2-2x-21}{x(x^3+7x)}}$
Here, in g (x), there are two factors, one linear and one quadratic. So, the expression can be decomposed as:
${\displaystyle \frac{x^2-2x-21}{x(x^3+7x)}=\frac{A}{x}+\frac{Bx+C}{x^2+7}}$
Here, Least Common Divisor is ${\displaystyle x(x^2+7)}$
Multiply both sides of (1) by the Least Common Divisor to clear fractions:
${\displaystyle x(x^2+7)\left[\frac{x^2-2x-21}{x(x^2+7x)}\right]=x(x^2+7)[\frac{-A}{x}+\frac{Bx+C}{x^2+7x}]}$
simplify to obtain:
${\displaystyle x^2+2x-21=A(x^2+7)+(Bx+C)x}$
${\displaystyle x^2+2x-21=A{x}^2+7+B{x}^2+Cx}$
${\displaystyle x^2+2x-21=(A+C)x^2+Cx+7A}$
Compare the coefficients of x,x? and constant terms:
${\displaystyle A+B=1,C=-2\text{nd7A=-2}nd7A=-21}$
Solve for A the equation ${\displaystyle 7A=—21}$
${\displaystyle 7A=-21}$
${\displaystyle A=-3}$
Substitute -3 for A in ${\displaystyle A+B=1}$ and simplify for B:
${\displaystyle (-3)+B=1}$
${\displaystyle B=4}$
Substitute the obtained values of A, B and Cin (1):