Question

How do you write the partial fraction decomposition of the rational expression ?

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

Answer 1

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

Explanation:

Note that `x^2+x+1 = (x+1/2)^2+3/4 > 0` for all real values of `x`, so has no linear factors with real coefficients.

So assuming that we want real coefficients in our partial fraction decomposition, we are looking for a decomposition of the form:

`(x^2+2x+2)/((x^2+x+1)(x+1)) = (Ax+B)/(x^2+x+1)+C/(x+1)`

Multiplying both sides by `(x^2+x+1)(x+1)` this becomes:

`x^2+2x+2 = (Ax+B)(x+1)+C(x^2+x+1)`

`color(white)(x^2+2x+2) = (A+C)x^2+(A+B+C)x+(B+C)`

Putting `x=-1`, we find:

`1 = (color(blue)(-1))^2+2(color(blue)(-1))+2 = C((color(blue)(-1))^2+(color(blue)(-1))+1) = C`

Equating the coefficient of `x^2`, we find:

`1 = A+C = A+1`

and hence `A=0`

Equating the constant term, we find:

`2 = B+C = B+1`

and hence `B=1`.

So:

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

Now we know the answer, we can demonstrate it more simply:

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