How do you write the partial fraction decomposition of the rational expression $\frac{1}{(x + 6) (x^{2} + 3)}$ $1/((x+6)(x^2+3))$ ?

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How do you write the partial fraction decomposition of the rational expression $\frac{1}{(x + 6) (x^{2} + 3)}$ $1/((x+6)(x^2+3))$ ?

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Answer 1 · 2018-03-25T07:53:18

$\frac{1}{(x + 6) (x^{2} + 3)} = \frac{1}{39 (x + 6)} - \frac{x - 6}{39 (x^{2} + 3)}$ $1/((x+6)(x^2+3)) = 1/(39(x+6))-(x-6)/(39(x^2+3))$

Explanation:

Given:

$\frac{1}{(x + 6) (x^{2} + 3)}$ $1/((x+6)(x^2+3))$

Note that $x^{2} + 3$ $x^2+3$ has no linear factors with real coefficients.

So, assuming we want a decomposition with real coefficients, it takes the form:

$\frac{1}{(x + 6) (x^{2} + 3)} = \frac{A}{x + 6} + \frac{B x + C}{x^{2} + 3}$ $1/((x+6)(x^2+3)) = A/(x+6)+(Bx+C)/(x^2+3)$

Multiplying both sides by $(x + 6) (x^{2} + 3)$ $(x+6)(x^2+3)$ this becomes:

$1 = A (x^{2} + 3) + (B x + C) (x + 6)$ $1 = A(x^2+3)+(Bx+C)(x+6)$

$1 = (A + B) x^{2} + (6 B + C) x + (3 A + 6 C)$ $color(white)(1) = (A+B)x^2+(6B+C)x+(3A+6C)$

Putting $x = - 6$ $x=-6$ we get:

$1 = A ({(- 6)}^{2} + 3) = 39 A$ $1 = A((color(blue)(-6))^2+3) = 39A$

So:

$A = \frac{1}{39}$ $A=1/39$

Then from the coefficient of $x^{2}$ $x^2$ , we have:

$A + B = 0$ $A+B = 0$

So:

$B = - A = - \frac{1}{39}$ $B = -A = -1/39$

From the constant term we have:

$1 = 3 A + 6 C$ $1 = 3A+6C$

Hence:

$C = \frac{1}{6} (1 - 3 A) = \frac{1}{6} (1 - \frac{1}{13}) = \frac{2}{13} = \frac{6}{39}$ $C = 1/6(1-3A) = 1/6(1-1/13) = 2/13 = 6/39$

So:

$\frac{1}{(x + 6) (x^{2} + 3)} = \frac{1}{39 (x + 6)} - \frac{x - 6}{39 (x^{2} + 3)}$ $1/((x+6)(x^2+3)) = 1/(39(x+6))-(x-6)/(39(x^2+3))$

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How do you write the partial fraction decomposition of the rational expression $\frac{1}{(x + 6) (x^{2} + 3)}$ $1/((x+6)(x^2+3))$ ?

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Explanation:

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How do you write the partial fraction decomposition of the rational expression 1(x+6)(x2+3) 1/((x+6)(x^2+3))?

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How do you write the partial fraction decomposition of the rational expression $\frac{1}{(x + 6) (x^{2} + 3)}$ $1/((x+6)(x^2+3))$ ?