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x/(x^2+4)^2 Partial Fractions?
How do I do it?
2 Answers
- 7 years agoFavorite Answer
I couldn't get any breakdown...but here's what I did:
Ax+B over (x^2+4) + Cx+D over (x^2+4)^2
working only in the numerators....
(Ax+B)(x^2+4) +Cx+D has to equal 1x from the original numerator.
multiply out:
Ax^3 +4Ax +Bx^2 +4B +Cx +D =1x
or
A=0 (there are no x^3 terms), B=0 (there are no x^2 terms) and D=0 (there are no constants)
leaving Cx=1x or C=1 which is the original fraction...
so sorry, I'm stuck here!
- JuanLv 67 years ago
x/(x^2+4)^2
that is easy
[(A*x + B)/(4 + x^2)] + [(C*x + D)/(4 + x^2)^2] = x/(4 + x^2)^2
(A*x + B)*(4 + x^2) + (C*x + D) = x
4x*A + (x^3)*A + 4*B + (x^2)*B + C*x + D - x = 0
+ 4*B + + D = 0
(x^3)*A + (x^2)*B + x*(C + 4A - 1) + 4*B + D = 0
A = 0
B = 0
C + 4*A - 1 = 0
4*B + D = 0
A = 0; B = 0; C = 1; D = 0
This means you their is no partial fractions as
[(x + 0)/(4 + x^2)^2] = x/(4 + x^2)^2
