Answer:
Partial fraction = 1/(x-2) - x/(x^2+1)
Step-by-step explanation:
Question:
Express 2x+1/(x-2)(x²+1) as a partial fraction.
Note: it will be assumed that there was a typo in the interpretation of parentheses to mean
(2x+1) / ( (x-2)(x^2+1) )
Let
(2x+1) / ( (x-2)(x^2+1) ) = A/(x-2) + (Bx+C)/(x^2+1) .........................(0)
(2x+1) / ( (x-2)(x^2+1) ) = (A(x^2+1)+(Bx+C)(x-2)) / ( (x-2)(B/(x^2+1) )
(2x+1) / ( (x-2)(x^2+1) ) = (Ax^2+A+Bx^2+(C-2B)x-2C) / ( (x-2)(B/(x^2+1) )
(2x+1) / ( (x-2)(x^2+1) ) = ( (A+B)x^2+(C-2B)x+A-2C ) / ( (x-2)(B/(x^2+1) )
Match numerators
2x+1 = (A+B)x^2+(C-2B)x+A-2C
Match coefficients,
A+B = 0 ..................(1)
-2B+C = 2 .................(2)
A-2C = 1 ...................(3)
Solve for A, B and C
Substitute A from (1) in (3)
-B - 2C =1
transpose and solve for B
B = -2C-1 ....................(4)
Substitue B from (4) in (2)
-2(-2C-1) + C = 2
simplify
5C = 2-2 = 0
C=0 ..........................(5)
substitute (5) in (4)
B = -2C-1 = -1 ...............(6)
Substitue (6) in (1)
A+(-1) = 0
A=1 .............................(7)
Using values from (7), (6) and (5) to substitute in (0)
we get
(2x+1) / ( (x-2)(x^2+1) ) = 1/(x-2) - x/(x^2+1)
as the required partial fraction
Answer:
[tex]\dfrac{1}{x-2}-\dfrac{x}{x^2+1}[/tex]
Step-by-step explanation:
The partial fraction form will be ...
[tex]\dfrac{2x+1}{(x-2)(x^2+1)}=\dfrac{A}{x-2}+\dfrac{Bx+C}{x^2+1}\\\\=\dfrac{A(x^2+1)+(x-2)(Bx+C)}{(x-2)(x^2+1)}\\\\\text{Equating numerators, we have ...}\\\\2x+1=(A+B)x^2+(C-2B)x+(A-2C)\\\\A=-B;\ C+2A=2;\ A-2C=1\\\\C+2(2C+1)=2\quad\text{substitute for A}\\\\5C+2=2\ \rightarrow\ C=0;\ A=1;\ B=-1\\\\\text{So the expansion is ...}\\\\\dfrac{2x+1}{(x-2)(x^2+1)}=\boxed{\dfrac{1}{x-2}-\dfrac{x}{x^2+1}}[/tex]
