Question

Express 2x+1/(x-2)(x²+1) as a partial fraction.

Answer 1

Partial fraction = 1/(x-2) - x/(x^2+1)

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 2

`(1)/(x-2)-(x)/(x^2+1)`

Explanation:

The partial fraction form will be ...

`(2x+1)/((x-2)(x^2+1))=(A)/(x-2)+(Bx+C)/(x^2+1)\\\\=(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 ...}\\\\(2x+1)/((x-2)(x^2+1))=\boxed{(1)/(x-2)-(x)/(x^2+1)}`