Chapter 11, Problem 25CT

To determine

The partial fraction decomposition of the rational expression $\frac{4x^2-3}{x{\left(x^2+3\right)}^2}$.

Answer to Problem 25CT

Solution:

The partial fraction decomposition of the rational expression $\frac{4x^2-3}{x{\left(x^2+3\right)}^2}$ is $\frac{4x^2-3}{x{\left(x^2+3\right)}^2}=\frac{-\frac{1}{3}}{x}+\frac{\frac{1}{3}x}{x^2+3}+\frac{5x}{{\left(x^2+3\right)}^2}$.

Explanation of Solution

Given information:

The rational expression, $\frac{4x^2-3}{x{\left(x^2+3\right)}^2}$.

Explanation:

Here denominator is $x{\left(x^2+3\right)}^2$.

Thus, the denominator has a linear factor $x$ and repeated irreduciblequadratic factor $x^2+3$.

Therefore, partial decomposition of $\frac{4x^2-3}{x{\left(x^2+3\right)}^2}$ takes form as $\frac{A}{x}+\frac{Bx+C}{x^2+3}+\frac{Dx+E}{{\left(x^2+3\right)}^2}$.

Then, $\frac{4x^2-3}{x{\left(x^2+3\right)}^2}=\frac{A}{x}+\frac{Bx+C}{x^2+3}+\frac{Dx+E}{{\left(x^2+3\right)}^2}$ where $A,B,C,D$ and $E$ are to be determined.

Multiply both sides of equation by $x{\left(x^2+3\right)}^2$.

$\Rightarrow 4x^2-3=A{\left(x^2+3\right)}^2+\left(Bx+C\right)x\left(x^2+3\right)+\left(Dx+E\right)x$

By simplifying,

$\Rightarrow 4x^2-3=A\left(x^4+6x^2+9\right)+\left(Bx+C\right)\left(x^3+3x^2\right)+D{x}^2+Ex$

$\Rightarrow 4x^2-3=A{x}^4+6A{x}^2+9A+B{x}^4+3B{x}^2+C{x}^3+3Cx+D{x}^2+Ex$

Combine the like terms,

$\Rightarrow 4x^2-3=\left(A+B\right)x^4+C{x}^3+\left(6A+3B+D\right)x^2+\left(3C+E\right)x+9A$

By equating coefficients of $x$ and constants,

$A+B=0$ ……. (1)

$C=0$……. (2)

$6A+3B+D=4$ … (3)

$3C+E=0$ … (4)

$9A=-3$ … (5)

From (5),

$\Rightarrow A=\frac{-3}{9}=-\frac{1}{3}$

Substitute $A=-\frac{1}{3}$ in equation (1),

$\Rightarrow -\frac{1}{3}+B=0$

$\Rightarrow B=\frac{1}{3}$

From equation (2),

$\Rightarrow C=0$

Substitute $A=-\frac{1}{3},B=\frac{1}{3}$ in equation (3)

$\Rightarrow 6\left(-\frac{1}{3}\right)+3\left(\frac{1}{3}\right)+D=4$

$\Rightarrow -2+1+D=4$

$\Rightarrow -1+D=4$

$\Rightarrow D=5$

From equation (4), $3C+E=0$

$\Rightarrow 3\left(0\right)+E=0$

$\Rightarrow E=0$

Thus, $A=-\frac{1}{3},B=\frac{1}{3},C=0,D=5,E=0$

Hence, $\frac{4x^2-3}{x{\left(x^2+3\right)}^2}=\frac{-\frac{1}{3}}{x}+\frac{\frac{1}{3}x}{x^2+3}+\frac{5x}{{\left(x^2+3\right)}^2}$.