Today, we continue to look at integrals using the different integration techniques shown in Calculus 2. In today’s example we find \(\int \frac{x^{2}-18}{x^{3}-2x^{2}+x}dx\). After determining that the best technique of integration to do this is by using reduction formulas, we then evaluate this integral.

Hopefully you will join me each day as I go through these integrals in both a post and in a video. Along the way, I hope you learn something, and, if you are taking Calculus 2, you are in a great position to do very well on any upcoming exams you may have. You can find more Calculus 2 study help posts here. I also have a decision tree available so that you can follow along as we determine what technique of integration to use. I also have a useful list of formula sheet available.

**Find \(\int \frac{x^{2}-18}{x^{3}-2x^{2}+x}dx\).**When finding \(\int \frac{x^{2}-18}{x^{3}-2x^{2}+x}dx\), we begin by going through the questions in our decision tree to determine the best technique of integration to use.

- Can you guess the answer?

I will work under the assumption that we are not able to guess integral. - Is there an inside function?

Here we do have an inside function. This function would be the bottom \(x^{3}-2x^{2}+x\), so we let this be \(i\).

By letting \(i=x^{3}-2x^{2}+x\), we find that \(di=3x^{2}-4x+1dx\), so \(dx=\frac{di}{3x^{2}-4x+1}\). If we substitute this back into the equation we get that

\begin{align*}

\int \frac{x^{2}-18}{x^{3}-2x^{2}+x}dx=\int \frac{x^{2}-18}{i}\frac{di}{3x^{2}-4x+1}.

\end{align*}

After finishing our substitution, we notice that we are not able to cancel out the \(x\)s. Furthermore, the remaining \(x\)s do no look like \(i\) enough to allow for some changing of terms into \(i\)s. We therefore start the process over, choosing that there is no inside function.

**Integration by parts.**

In our second run through the decision tree we get

- Can you guess the answer?

I will work under the assumption that we are not able to guess integral. - Is there an inside function?

No, since we have seen this won’t work out. - Is there a product of functions?

There is if we look at this as \((x^{2}-18)*\frac{1}{x^{3}-2x^{2}+x}\) - Is this a product of trig functions?

No.

We therefore continue using integration by parts. In this case, if we let \(S’=\frac{1}{x^{3}-3x^{2}+x}\), we will have to integrate this piece. Looking at this, it’s not immediately clear how to do this. Furthermore, if we could, it would likely result in a more complicated function inside a new integral which wouldn’t help us. We would then choose \(S’=x^{2}-18\) and \(F=\frac{1}{x^{3}-3x^{2}+x}\). However, upon integrating \(S’\) and differentiating \(F\), we can note fairly quickly that \(\int SF’dx\) will again be more complicated than the original integral. Therefore, we should start over, skipping integration by parts.

**Partial Fraction Decomposition.**

Going through the decision tree again, we find the following.

- Can you guess the answer?

I will work under the assumption that we are not able to guess integral. - Is there an inside function?

No, since we have seen this won’t work out. - Is there a product of functions?

No, since we have seen this won’t work out. - Is there a rational function?

Yes, this is a rational function.

We therefore try to break the function into smaller pieces using partial fraction decomposition. When doing this, I will leave the integral until we are finished and attempt to simplify

\begin{align*}

\frac{x^{2}-18}{x^{3}-2x^{2}+x}.

\end{align*}

The first step in simplifying this is to factor the bottom. We note quickly that \(x\) is a common factor, so we find \(x^{3}-2x^{2}+x=x(x^{2}-2x+1)\). Looking at the quadratic portion, we note that this will factor into \((x-1)^{2}\), so we get \(x^{3}-2x^{2}+x=x(x-1)^{2}\). Now that we have factored the bottom, we know that we can factor the original into pieces were the new denominator are the terms in the original. However, since the \((x-1)\) has multiplicity 2, we need to have two fractions for this term, with the result being, for some \(A,B\) and \(C\) constant,

\begin{align*}

\frac{x^{2}-18}{x^{3}-2x^{2}+x}&=\frac{A}{x}+\frac{B}{(x-1)^{2}}+\frac{C}{x-1}.

\end{align*}

You can find the general form for the decomposition in the integral formula sheet.

In order to find \(A,B\) and \(C\), we will first multiply by the denominator on the left and get

\begin{align*}

x^{2}-18&=A(x-1)^{2}+Bx+Cx(x-1).

\end{align*}

Now, we could distribute the right out and create a system of equations. However, we know that if we are to have a solution, it must work for all \(x\)s. Therefore, we can limit what \(A,B\) and \(C\) are by choosing strategic \(x\)s. That is, if we let \(x=0\) we find that

\begin{align*}

-18&=A(-1)^{2}+0 \\

-18&=A.

\end{align*}

If we now let \(x=1\), we get that

\begin{align*}

1-18&=0+B+0 \\

-17&=B.

\end{align*}

We now must choose any other \(x\), so let \(x=2\). Using the \(A\) and \(B\) we have already found, we get

\begin{align*}

4-18&=-18(2-1)^{2}-17(2)+C(2)(2-1) \\

-14 &=-18-34+2C \\

38&=2C \\

19&=C.

\end{align*}

Therefore, we have that

\begin{align*}

\frac{x^{2}-18}{x^{3}-2x^{2}+x}&=\frac{-18}{x}+\frac{-17}{(x-1)^{2}}+\frac{19}{x-1}.

\end{align*}

We can now find the integral as

\begin{align*}

\int \frac{x^{2}-18}{x^{3}-2x^{2}+x}dx&=-18 \int \frac{1}{x}dx-17 \int \frac{1}{(x-1)^{2}}dx+19 \int \frac{1}{x-1}dx \\

&=-18 \ln|x|+17(x-1)^{-1}+19\ln|x-1|+c.

\end{align*}

In this past integral we work under the assumption you could guess the integrals in order to save some. If you can’t, that is expected. However, I will refer you to Integration Techniques 1-Substitution for an example using integration by substitution. If of these pieces can be completed using that technique.

**Conclusion**

In this problem, we noted that we did have to try a few integration techniques before we finally realized that we would have to simplify the problem using partial fraction decomposition before integrating. While it did take some time to try these techniques, the attempts went quickly enough that they were worth looking at because, if they had worked, they would have been easier than using partial fraction decomposition.

I hope you learned something from this post and found it helpful. If you did, make sure to share it and the video with anyone else you know that may need some extra help with integrals.