(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...
(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
oops. forgot the 2x
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
To evaluate the integral I=∫ f(x) dx using the information from parts (a) to (c), follow these steps:
1. Rewrite the function f(x) as a polynomial plus a proper rational function using the result from part (a):
f(x) = (x+2) + (3x^2-2x-3)/(x^3-x^2+x-1)
2. Split the integral into two parts:
I = ∫ [(x+2) + (3x^2-2x-3)/(x^3-x^2+x-1)] dx
3. Integrate the polynomial part:
∫ (x+2) dx = (1/2)x^2 + 2x + C₁, where C₁ is the constant of integration.
4. Integrate the proper rational function part using partial fractions. Let's work on the fraction (3x^2-2x-3)/(x^3-x^2+x-1) step by step:
- Factor the denominator:
x^3 - x^2 + x - 1 = (x^2 + 1)(x - 1)
- Express the proper rational function as a sum of partial fractions:
(3x^2-2x-3)/(x^3-x^2+x-1) = A/(x-1) + (Bx + C)/(x^2+1)
- Multiply both sides by the common denominator (x^3 - x^2 + x - 1):
3x^2 - 2x - 3 = A(x^2 + 1) + (Bx + C)(x - 1)
- Equate the coefficients of corresponding powers of x:
3x^2 - 2x - 3 = Ax^2 + A + Bx^2 - Bx + Cx - C
- Group the like terms:
(A + B)x^2 + (C - B)x + (A - C) = 3x^2 - 2x - 3
- Equate the coefficients to solve for A, B, and C:
A + B = 3
C - B = -2
A - C = -3
- Solve this system of linear equations to find the values of A, B, and C. Once you find those values, substitute them back into the partial fraction decomposition.
5. Integrate the resulting partial fractions:
∫ [A/(x-1) + (Bx + C)/(x^2+1)] dx
- Integrate A/(x-1) using the natural logarithm:
∫ A/(x-1) dx = A ln |x - 1| + C₂, where C₂ is the constant of integration.
- Integrate (Bx + C)/(x^2 + 1) using arctangent:
∫ (Bx + C)/(x^2 + 1) dx = B/2 ln |x^2 + 1| + C/√2 arctan(x) + C₃, where C₃ is the constant of integration.
6. Combine the integrated parts and the constants of integration to obtain the final result for the integral I.
That's how you would use the information from parts (a) to (c) to evaluate the integral. Remember to substitute the values of A, B, and C obtained from solving the partial fractions before integrating.