(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-1x + C
(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...
3 answers
oops. forgot the 2x
But I'm sure you caught that.
But I'm sure you caught that.
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
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