I get
(27x+14)/(x^2+1) - (27/25)/(x-2) + (8/5)/(x-2)^2
Resolve in partial fractions (x+6)/((x^2+1)(x-2)^2)
Using let x = ...
3 answers
how did you work this out please ?
Review your text on partial fractions. You know that the sum will be
(Ax+B)/(x^2+1) + C/(x-2) + D/(x-2)^2
If you put all that over a common denominator, you get
(Ax+B)(x-2)^2 + C(x-2)(x^2+1) + D(x^2+1)
------------------------------------------------
(x^2+1)(x-2)^2
Expand everything in the numerator and you get
(A+C)x^3 + (-4A+B-2C+D)x^2 + (4A-4B+C)x + (4B-2C+D)
------------------------------------
(x^2+1)(x-2)^2
If that is identical to (x+6)/((x^2+1)(x-2)^2)), then we must have all the coefficients in the two numerators be identical. That is,
A+C=0
-4A+B-2C+D = 0
4A-4B+C = 1
4B-2C+D = -6
If I haven't made any typos, if you solve that system you get the numbers I presented earlier.
(Ax+B)/(x^2+1) + C/(x-2) + D/(x-2)^2
If you put all that over a common denominator, you get
(Ax+B)(x-2)^2 + C(x-2)(x^2+1) + D(x^2+1)
------------------------------------------------
(x^2+1)(x-2)^2
Expand everything in the numerator and you get
(A+C)x^3 + (-4A+B-2C+D)x^2 + (4A-4B+C)x + (4B-2C+D)
------------------------------------
(x^2+1)(x-2)^2
If that is identical to (x+6)/((x^2+1)(x-2)^2)), then we must have all the coefficients in the two numerators be identical. That is,
A+C=0
-4A+B-2C+D = 0
4A-4B+C = 1
4B-2C+D = -6
If I haven't made any typos, if you solve that system you get the numbers I presented earlier.
2 answers