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.

x^2-x+6

----------
x (x^2-3)
well the bottom is a bore since I can not really factor it ((x-sqrt 3)(x+sqrt 3) is no fun) so I will say
a/(x) + (bx+c)/(x^2-3)
then
a(x^2-3) +(bx+c)(x) = 1 x^2 - 1 x + 6
a x^2 - 3a + b x^2 + c x = 1 x^2 - 1 x + 6
a + b = 1
c = -1
-3 a = 6
so
a = -2
b = 3
c = -1
and
-2/x + (3x-1)/(x^2-3) is what we have to integrate
this is
-2 integral dx/x + 3 integral x dx/(x^2-3) - integral (dx/(x^2-3)

f(x) = (x^2-x+6)/(x^3-3x)

Factor the numerator:

x^3 - 3x = x(x^2 - 3) =

x[x-sqrt(3)][x+sqrt(3)]

And it follows that the function f(x) must be of the form:

f(x) = A/x + B/(x-sqrt(3)) +
C/(x+sqrt(3))

To find A multiply both sides by x and take the limit x --->0:

A = -2

To find B multiply both sides by
(x-sqrt(3) ) and take the limit
x --->sqrt(3):

B = 3/2 - 1/6 sqrt(3)

To find C multiply both sides by
(x+sqrt(3) ) and take the limit
x ---> -sqrt(3):

C = 3/2 + 1/6 sqrt(3)

To find the integral of the given rational function, you need to use the method of partial fractions. The process involves decomposing the rational function into simpler fractions and then integrating each term individually.

Step 1: Factorize the denominator:
Start by factoring the denominator, x^3 - 3x. It can be factored as x(x^2 - 3).

Step 2: Write the partial fraction decomposition:
Write the rational function as the sum of two fractions with unknown numerators, A/x + (Bx + C)/(x^2 - 3).

Step 3: Clear the fractions:
Multiply through by the denominator, x(x^2 - 3), to eliminate the fractions:
x^2 - x + 6 = A(x^2 - 3) + (Bx + C)x

Step 4: Equate coefficients:
Compare the coefficients of the corresponding powers of x on both sides to establish equations:
For the constant term: 6 = -3A (coefficient of x^0)
For the linear term: -1 = B + C (coefficient of x^1)
For the quadratic term: 1 = A (coefficient of x^2)

Step 5: Solve for unknowns:
Solve the system of equations to find the values of A, B, and C. From the first equation, A = -2. From the second equation, B = -1 - C. Substitute the values of A and B into the third equation to find C.

Step 6: Rewrite the original integral:
Rewrite the original integral as the sum of two separate integrals:
∫(x^2 - x + 6)/(x^3 - 3x) dx = -2∫(1/x) dx + ∫((Cx - 1)/(x^2 - 3)) dx

Step 7: Integrate each term separately:
The integral of 1/x, ∫(1/x) dx, is the natural logarithm (ln) of the absolute value of x:
-2∫(1/x) dx = -2ln|x| + C1, where C1 is the constant of integration.

For the second term, you can use trigonometric substitution or complete the square to integrate it. The result involves the inverse tangent (arctan) function:
∫((Cx - 1)/(x^2 - 3)) dx = (C/2)ln|(x^2 - 3)| - (1/2√3)arctan(x/√3) + C2, where C2 is another constant of integration.

Step 8: Combine the results:
Finally, combine the two integrals by adding the constants of integration to get the complete answer.

Therefore, the integral of (x^2 - x + 6)/(x^3 - 3x) includes ln and arctan terms, resulting from the partial fraction decomposition.