How do you write the partial fraction decomposition of the rational expression 1/((x+6)(x^2+3))1(x+6)(x2+3)?
1 Answer
Explanation:
Given:
1/((x+6)(x^2+3))1(x+6)(x2+3)
Note that
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)1(x+6)(x2+3)=Ax+6+Bx+Cx2+3
Multiplying both sides by
1 = A(x^2+3)+(Bx+C)(x+6)1=A(x2+3)+(Bx+C)(x+6)
color(white)(1) = (A+B)x^2+(6B+C)x+(3A+6C)1=(A+B)x2+(6B+C)x+(3A+6C)
Putting
1 = A((color(blue)(-6))^2+3) = 39A1=A((−6)2+3)=39A
So:
A=1/39A=139
Then from the coefficient of
A+B = 0A+B=0
So:
B = -A = -1/39B=−A=−139
From the constant term we have:
1 = 3A+6C1=3A+6C
Hence:
C = 1/6(1-3A) = 1/6(1-1/13) = 2/13 = 6/39C=16(1−3A)=16(1−113)=213=639
So:
1/((x+6)(x^2+3)) = 1/(39(x+6))-(x-6)/(39(x^2+3))1(x+6)(x2+3)=139(x+6)−x−639(x2+3)
