# Integral with partial fractions

## 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. :)

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I'm pretty sure you lost the sign on your A.

Are you sure? I just pulled it out in front of the whole fraction when I substituted the numerical values in.

$$-\frac{1}{x}+9\int[\frac{C}{x-3}+\frac{A}{x}+\frac{B}{x^2}]dx$$
$$-\frac{1}{x}+9\int[\frac{\frac{1}{9}}{x-3}-\frac{\frac{1}{9}}{x}-\frac{\frac{1}{3}}{x^2}]dx$$

Mark44
Mentor
@QuantumCurt, I haven't checked your work... When you use decomposition by partial fractions, it's a good idea to check your work by adding the (in this case, 3) fractions to make sure you end up with what you started.

• QuantumCurt
Sorry, I missed that you switched the order of A and C.

@Mark44 , that's good advice. I hadn't added the decomposition back together to check that it came out the same. I just checked it, and it does indeed come out the same.

Mark44
Mentor
OK, then since the numbers check, you can check your integration by taking the derivative of your answer. With a bit of albebraic manipulation, you should be able to get back to the original integrand.

• QuantumCurt
Also good advice. I just took the derivative of my answer, and after combining the fractions I ended up with my original function.

So...am I right in thinking that Wolfram might simply be getting this one wrong?

Dick
Homework Helper
Also good advice. I just took the derivative of my answer, and after combining the fractions I ended up with my original function.

So...am I right in thinking that Wolfram might simply be getting this one wrong?
You are both somewhat right. The derivative of log(3-x) is the same as the derivative of log(x-3). The usual thing is to write the integral of 1/(x-3) as log(|x-3|).

• QuantumCurt