what is the integral of x^2-x+6/x^3-3x?
the process involves partial fractions, and the answer is supposed to include ln and arctan... i just don't know how to get there.
Please use parentheses to clarify what the numerator and denominator are in that function
To solve this integral using partial fractions, we can follow these steps:
Step 1: Factorize the denominator:
First, factorize the denominator, x^3 - 3x. It can be written as x(x^2 - 3).
Step 2: Express the fraction in partial fractions:
The given expression can be written as:
(x^2 - x + 6) / [x(x^2 - 3)]
Since the degree of the numerator is less than the degree of the denominator, we can divide the numerator by the denominator using long division:
x - 3 (quotient)
-------------
x(x^2 - 3) │ x^2 - x + 6
- (x^2 - 3x)
2x + 6
- (2x - 6)
12
The remainder is 12, so we can rewrite the expression as:
x - 3 + (12 / [x(x^2 - 3)])
Step 3: Express the fraction using partial fractions:
Now, we need to express the fraction 12 / [x(x^2 - 3)] in partial fractions.
1. Decompose the denominator:
x(x^2 - 3) can be rewritten as x(x - sqrt(3))(x + sqrt(3)) since x^2 - 3 can be factored as (x - sqrt(3))(x + sqrt(3)).
2. Express the fraction using partial fractions:
12 / [x(x - sqrt(3))(x + sqrt(3))] can be rewritten as A/x + B/(x - sqrt(3)) + C/(x + sqrt(3)).
To find A, B, and C, we need to find a common denominator:
12 = A(x - sqrt(3))(x + sqrt(3)) + B*x(x + sqrt(3)) + C*x(x - sqrt(3))
Expand and collect like terms:
12 = A(x^2 - 3) + B(x^2 + sqrt(3)x) + C(x^2 - sqrt(3)x)
Now, we can equate the coefficients of the like terms:
For the terms with x^2:
0 = A + B + C
For the terms with x^1:
0 = -3A + sqrt(3)B - sqrt(3)C
For the constant terms:
12 = -3A
Solve the equations simultaneously to find the values of A, B, and C.
From the first equation, we have A = -4.
Plugging A = -4 in the second equation, we get -3(-4) + sqrt(3)B - sqrt(3)C = 0.
Simplifying, we get sqrt(3)B - sqrt(3)C = 12.
Dividing the above equation by sqrt(3), we have B - C = 4.
From the third equation, we have -3A = 12, which simplifies to A = -4.
Substituting A = -4 in the first equation, we get -4 + B + C = 0.
Simplifying, we have B + C = 4.
From the equations B - C = 4 and B + C = 4, we can solve for B and C to find that B = 2 and C = 2.
Therefore, the fraction 12 / [x(x^2 - 3)] can be rewritten as:
(-4/x) + (2 / (x - sqrt(3))) + (2 / (x + sqrt(3)))
Step 4: Integrate each term using basic integration formulas:
Using the partial fraction decomposition from Step 3, the original integral can be written as:
∫ [(x - 3) - (4/x) + (2 / (x - sqrt(3))) + (2 / (x + sqrt(3))))] dx
Integrating each term:
∫ (x - 3) dx - ∫ (4/x) dx + ∫ (2 / (x - sqrt(3))) dx + ∫ (2 / (x + sqrt(3)))) dx
First term: ∫ (x - 3) dx = (x^2 / 2) - 3x + C1
Second term: ∫ (4 / x) dx = 4 ln |x| + C2
Third term: ∫ (2 / (x - sqrt(3))) dx = 2 ln |x - sqrt(3)| + C3
Fourth term: ∫ (2 / (x + sqrt(3)))) dx = 2 ln |x + sqrt(3)| + C4
where C1, C2, C3, and C4 are constants of integration.
Therefore, the integral of (x^2 - x + 6) / (x^3 - 3x) is:
(x^2 / 2) - 3x + 4 ln |x| + 2 ln |x - sqrt(3)| + 2 ln |x + sqrt(3)| + C
where C is the constant of integration.