Question

(a) Consider the improper function f(x)= (x^4+x^3+2x^2+x-1)/(x^3-x^2+x-1). Use long division to convert f(x) into a polynomial plus a proper rational function.
-->I ended up with : (x+2) + (3x^2-2x-3)/(x^3-x^2+x-1)

(b) Factor the polynomial Q(x) = x^3 - x^2 + x -1 into a linear factor and a quadratic factor. (Hint:Try evaluating f(x) for small integers x.)
--> I don't understand the hint so what I have so far is:
x^3 - x^2 + x -1=0
x^3 - x^2 + x = 1
x(x^2-x+1)=1

(c) Use partial fractions to expand the rational function g(x) = (3x^2 +1)/(x-1)(x^2+1)
--> g(x)= A/x-1 + B/x^2+1
3x^2+1 = A(x^2 +1) + B(x-1)
3x^2+1 = Ax^2 + A + Bx -B
Unsure what to do after this

(d) Using parts (a) to (c) above, evaluate the integral I=∫ f(x) dx
--> how would I use (c) for this part of the question...

Answer 1

(a) I got (x+2) + (3x^2+1)/(x^3-x^2+x-1)
(b) x^3 - x^2 + x -1 = x^2(x-1) + (x-1) = (x^2+1)(x-1)
(c) You forgot how to handle quadratic denominators.
A/(x-1) + (Bx+C)/(x^2+1) = (A+B)x^2 + Cx + (A-B-C) = 3x^2+1
So the solution is A=2, B=1, C=1 and you have
(3x^2+1)/(x^3-x^2+x-1) = 2/(x-1) + (x+1)/(x^2+1)
(d)
∫ (x^4+x^3+2x^2+x-1)/(x^3-x^2+x-1) dx
= ∫ x+2 + 2/(x-1) + (x+1)/(x^2+1) dx
= ∫ x+2 + 2/(x-1) + (1/2) * (2x)/(x^2+1) + 1/(x^2+1) dx
= 1/2 x^2 + 2ln(x-1) + 1/2 ln(x^2+1) + tan^{^-1}x + C

Answer 2

oops. forgot the 2x
But I'm sure you caught that.

Answer 3

Hello,
Thank you very much for your help with my question. If it's possible, could you please explain why you multiply 1/2 with 2(x)/(x^2+1) in this part:
= ∫ x+2 + 2/(x-1) + (1/2) * (2x)/(x^2+1) + 1/(x^2+1) dx