Posted by **belinda** on Sunday, March 2, 2008 at 5:15pm.

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 dont know how to get there.

- calculus -
**Damon**, Sunday, March 2, 2008 at 5:51pm
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)

- calculus -
**Count Iblis**, Sunday, March 2, 2008 at 6:02pm
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)

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