How do I decompose the rational expression $\frac{x - 3}{x^{3} + 3 x}$ $(x-3)/(x^3+3x)$ into partial fractions?

Question

5368 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-17T12:39:56

The answer is $= - \frac{1}{x} + \frac{x + 1}{x^{2} + 3}$ $=-1/x+(x+1)/(x^2+3)$

Perform the decomposition into partial fractions after factorising the denominator

$\frac{x - 3}{x^{3} + 3 x} = \frac{x - 3}{x (x^{2} + 3)}$ $(x-3)/(x^3+3x)=(x-3)/(x(x^2+3))$

$= \frac{A}{x} + \frac{B x + C}{x^{2} + 3}$ $=A/x+(Bx+C)/(x^2+3)$

$= \frac{A (x^{2} + 3) + x (B x + C)}{x (x^{2} + 3)}$ $=(A(x^2+3)+x(Bx+C))/(x(x^2+3))$

The denominators are the same, compare the numerators

$x - 3 = A (x^{2} + 3) + x (B x + C)$ $x-3=A(x^2+3)+x(Bx+C)$

Let $x = 0$ $x=0$ , $\Rightarrow$ $=>$ , $- 3 = 3 A$ $-3=3A$ , $\Rightarrow$ $=>$ , $A = - 1$ $A=-1$

Coefficients of $x^{2}$ $x^2$

$0 = A + B$ $0=A+B$ , $\Rightarrow$ $=>$ , $B = - A = 1$ $B=-A=1$

Coefficients of $x$ $x$

$1 = C$ $1=C$

Therefore,

$\frac{x - 3}{x^{3} + 3 x} = - \frac{1}{x} + \frac{x + 1}{x^{2} + 3}$ $(x-3)/(x^3+3x)=-1/x+(x+1)/(x^2+3)$

How do I decompose the rational expression (x-3)/(x^3+3x)x−3x3+3x(x-3)/(x^3+3x) into partial fractions?