# Integral with partial fractions

1. Feb 27, 2015

### QuantumCurt

1. The problem statement, all variables and given/known data

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.

$$\int\frac{x+6}{x^2(x-3)}dx$$

3. 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. :)

2. Feb 27, 2015

### MostlyHarmless

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

3. Feb 27, 2015

### QuantumCurt

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

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

4. Feb 27, 2015

### Staff: 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.

5. Feb 27, 2015

### MostlyHarmless

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

6. Feb 27, 2015

### QuantumCurt

@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.

7. Feb 27, 2015

### Staff: 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.

8. Feb 27, 2015

### 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?

9. Feb 27, 2015

### Dick

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|).

10. Feb 27, 2015

### QuantumCurt

Now that you mention that...I recall encountering this issue with Wolfram before when functions have integrated to logarithms. It often seems to put it in the log(c-x) form rather than log(x-c).

11. Feb 27, 2015

### QuantumCurt

Thanks for the help everyone!

I just finished my retro calculus assignment, and that was the only one that I got hung up on. Good to know that I still remember all of this stuff from calculus 2.