what is the integral of (x^2-x+6)/(x^3-3x)?

Question

the process involves partial fractions, and the answer is supposed to include ln and arctan... i just don't know how to get there.

Answer 1

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)

Answer 2

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)