To rewrite 1/(x^3+x^2+x+1) as partial fractions, we first need to factorize the denominator

1/(x^3+x^2+x+1)

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

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

Let this be equal to (Ax +B) / (x^2 +1) + C / (x +1)

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

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

=> Ax^2 + Bx +Ax + B + Cx^2 + C = 1

=> Ax^2 + Cx^2 = 0 , Ax + Bx = 0 , B + C = 1

=> A + C = 0 , A + B = 0 and B + C = 1

=> C = -A , B = -A and B + C = 1

=> -A + -A = 1

=> A = -1/2

C = 1/2 and B = 1/2

So (Ax +B) / (x^2 +1) + C / (x +1)

=> (-1/2)x +(1/2) / (x^2 +1) + (1/2) / (x +1)

=> (1 - x)/ 2(x^2 +1) + 1/2(x +1)

**Therefore the required result is (1-x)/[2(x^2 +1)] +1/2*(x +1)**

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now