Posted by belinda on Sunday, March 2, 2008 at 5:15pm.
what is the integral of (x^2x+6)/(x^33x)?
the process involves partial fractions, and the answer is supposed to include ln and arctan... i just don't know how to get there.

calculus  Damon, Sunday, March 2, 2008 at 5:51pm
x^2x+6

x (x^23)
well the bottom is a bore since I can not really factor it ((xsqrt 3)(x+sqrt 3) is no fun) so I will say
a/(x) + (bx+c)/(x^23)
then
a(x^23) +(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 + (3x1)/(x^23) is what we have to integrate
this is
2 integral dx/x + 3 integral x dx/(x^23)  integral (dx/(x^23)

calculus  Count Iblis, Sunday, March 2, 2008 at 6:02pm
f(x) = (x^2x+6)/(x^33x)
Factor the numerator:
x^3  3x = x(x^2  3) =
x[xsqrt(3)][x+sqrt(3)]
And it follows that the function f(x) must be of the form:
f(x) = A/x + B/(xsqrt(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
(xsqrt(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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