How do you express $\frac{x}{(x - 1) (x^{2} + 4)}$ $x/((x-1)(x^2+4)$ in partial fractions?

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How do you express $\frac{x}{(x - 1) (x^{2} + 4)}$ $x/((x-1)(x^2+4)$ in partial fractions?

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Answer 1 · 2016-11-07T15:57:21

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

Explanation:

Let $\frac{x}{(x - 1) (x^{2} + 4)} = \frac{A}{x - 1} + \frac{B x + C}{x^{2} + 4}$ $x/((x-1)(x^2+4))=A/(x-1)+(Bx+C)/(x^2+4)$
$= \frac{A (x^{2} + 4) + (B x + C) (x - 1)}{(x - 1) (x^{2} + 4)}$ $=(A(x^2+4)+(Bx+C)(x-1))/((x-1)(x^2+4))$
So $x = A (x^{2} + 4) + (B x + C) (x - 1)$ $x=A(x^2+4)+(Bx+C)(x-1)$
if $x = 0$ $x=0$ $\Rightarrow$ $=>$ $0 = 4 A - C$ $0=4A-C$
coefficients of $x^{2}$ $x^2$ $\Rightarrow$ $=>$ $0 = A + B$ $0=A+B$
coefficents of $x$ $x$ $\Rightarrow$ $=>$ $1 = - B + C$ $1=-B+C$
Solving for $A, B, C$ $A,B, C$
$A = \frac{1}{5}$ $A=1/5$
$B = - \frac{1}{5}$ $B=-1/5$
$C = \frac{4}{5}$ $C=4/5$

$:.x/((x-1)(x^2+4))=1/(5(x-1))+(-x+4)/(5(x^2+4))$

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How do you express $\frac{x}{(x - 1) (x^{2} + 4)}$ $x/((x-1)(x^2+4)$ in partial fractions?

1 Answer

Explanation:

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How do you express $\frac{x}{(x - 1) (x^{2} + 4)}$ $x/((x-1)(x^2+4)$ in partial fractions?