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

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Answer 1 · 2018-03-25T07:53:18

$1/((x+6)(x^2+3)) = 1/(39(x+6))-(x-6)/(39(x^2+3))$

Given:

$1/((x+6)(x^2+3))$

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

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

$1/((x+6)(x^2+3)) = A/(x+6)+(Bx+C)/(x^2+3)$

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

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

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

Putting $x=-6$ we get:

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

So:

$A=1/39$

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

$A+B = 0$

So:

$B = -A = -1/39$

From the constant term we have:

$1 = 3A+6C$

Hence:

$C = 1/6(1-3A) = 1/6(1-1/13) = 2/13 = 6/39$

So:

$1/((x+6)(x^2+3)) = 1/(39(x+6))-(x-6)/(39(x^2+3))$

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