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Homework Statement
I'm currently in Calculus 3, and the professor gave us a "retro assignment" which is basically a bunch of tough integrals from Calculus 2. I think my process here is valid, but when I check my answer on Wolfram, they're getting a slightly different final answer.
$$\int\frac{x+6}{x^2(x-3)}dx$$
The Attempt at a Solution
Upon first examination, two possibilities popped out at me.
I could rewrite the function as a sum of two fractions
$$\int[\frac{x}{x^2(x-3)}+\frac{6}{x^2(x-3)}]dx$$
$$\int[\frac{1}{x(x-3)}+\frac{6}{x^2(x-3)}]dx$$
This works, but also involves two partial fraction decompositions. The second possibility that popped out at me was to rewrite the integral in a way that avoided one of the partial fraction decompositions.
$$\int\frac{x+6}{x^2(x-3)}dx$$
$$\int\frac{x-3+9}{x^2(x-3)}dx$$
$$\int[\frac{x-3}{x^2(x-3)}+\frac{9}{x^2(x-3)}]dx$$
$$\int[\frac{1}{x^2}+\frac{9}{x^2(x-3)}]dx$$
$$-\frac{1}{x}+9\int\frac{1}{x^2(x-3)}dx$$
Now I've eliminated one of the partial fraction decompositions. I still have to do a decomposition on the second term.
$$\frac{1}{x^2(x-3)}$$
$$\frac{1}{x^2(x-3)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{(x-3)}$$
Then I multiply by the common denominator
$$1=Ax(x-3)+B(x-3)+Cx^2$$
I can see right away that by letting ##x=0## I get
$$1=-3B \Rightarrow B=-\frac{1}{3}$$
Similarly, if I let ##x=3##, I get
$$1=9C \Rightarrow C=\frac{1}{9}$$
That gives me two of my constants, but there's no immediately apparent value to find ##A##, so I rewrite the decomposition as
$$1=Ax(x-3)-\frac{1}{3}(x-3)+\frac{1}{9}x^2$$
$$9=9Ax(x-3)-3(x-3)+x^2$$
Then I let ##x## be an arbitrary value of 1
$$9=9A(-2)-3(-2)+1$$
$$A=-\frac{1}{9}$$
Now I can rewrite my integral as
$$-\frac{1}{x}+9\int[\frac{C}{x-3}+\frac{A}{x}+\frac{B}{x^2}]dx$$
$$-\frac{1}{x}+9\int[\frac{\frac{1}{9}}{x-3}-\frac{\frac{1}{9}}{x}-\frac{\frac{1}{3}}{x^2}]dx$$
$$-\frac{1}{x}+\int[\frac{1}{x-3}-\frac{1}{x}-\frac{3}{x^2}]dx$$
$$-\frac{1}{x}+\ln(x-3)-\ln(x)+\frac{3}{x}+C$$
Which then simplifies to
$$\frac{2}{x}+\ln(\frac{x-3}{x})+C$$
Now, when I check my answer in Wolfram, it gives me a solution of
$$\frac{2}{x}+\ln(3-x)-\ln(x)+C$$
Which reduces to
$$\frac{2}{x}+\ln(\frac{3-x}{x})+C$$
This solution has ##3-x## inside of the logarithm instead of the ##x-3## that I obtained. Did I introduce some kind of error in the way I initially broke the function of by subtracting 3 and adding 9? I've looked it over several times now and I can't find any arithmetic mistakes anywhere. Any help would be very much appreciated. :)