# Thread: Finding indefinite integral with PCD method?

1. ## Finding indefinite integral with PCD method?

Ok, so the problem is take the indefinite integral of $\displaystyle x^4 / (x-1)^3$. The section the homework problem is in taught the Partial Fraction Decomposition method to rewrite an integral in a way that it can be integrated.
When I do it, I have $\displaystyle x^4 = A(x-1)^2 + B(x-1) + C$ so the only convenient x value is 1, so I was able to solve for c, because $\displaystyle c=1$, but if I use any X value per the rules to find the other two, well, I can't isolate one. For instance, I tried using $\displaystyle x=2$ which gave me $\displaystyle A + B = 15$ which doesn't help me much.

Any ideas on how to use PCD (or a more efficient method if there is one) to solve this integral? Thanks!

2. As it's written, you can't apply partial fraction descomposition directly. You have to make a long division 'cause the degree of the numerator is greater than the denominator one.

I dunno why you require to apply such method, since this can be killed with a simple substitution.

3. Originally Posted by Krizalid
As it's written, you can't apply partial fraction descomposition directly. You have to make a long division 'cause the degree of the numerator is greater than the denominator one.

I dunno why you require to apply such method, since this can be killed with a simple substitution.
Right, but when I do a u sub, I end up with $\displaystyle u = x-1$ and $\displaystyle du=dx$ but I need to switch the u sub around so $\displaystyle x = u + 1$ to plug in for the top variable and then the integral would be $\displaystyle (u+1)^4 / u^3 du$ which is still pretty ugly because you either have to multiply out the top part and divide each term by $\displaystyle u^3$ or use polynomial division. I was just wondering if there was an easier way to do it.

4. Originally Posted by emttim84
the integral would be $\displaystyle (u+1)^4 / u^3 du$ which is still pretty ugly
This is not ugly

I'd consider $\displaystyle (1+u)^n$ ugly for $\displaystyle n\ge5.$ You only need to expand that and integrate term by term. Is it too hard?

At least memorize some binomial powers, it helps.

5. Originally Posted by Krizalid
This is not ugly

I'd consider $\displaystyle (1+u)^n$ ugly for $\displaystyle n\ge5.$ You only need to expand that and integrate term by term. Is it too hard?

At least memorize some binomial powers, it helps.
Condescension isn't exactly useful to me (nor do I care about it), just FYI. And judging by the answer I get when I multiply it out and integrate it, I'd have a hard time settling for that answer without a solutions manual to check it against. Oh, and conveniently, it's an even-numbered problem so the solutions aren't in the solutions manual, obviously.

6. Originally Posted by emttim84
Condescension isn't exactly useful to me (nor do I care about it), just FYI. And judging by the answer I get when I multiply it out and integrate it, I'd have a hard time settling for that answer without a solutions manual to check it against. Oh, and conveniently, it's an even-numbered problem so the solutions aren't in the solutions manual, obviously.
Krizalid's reply was not condescending. For the record I think that the question asked, although perhaps a touch prevocative, was nevertheless reasonable ......

It looks to me like you're ticked off over a few other things, things that Krizalid is not responsible for.

fyi $\displaystyle \, (u + 1)^4 = u^4 + 4 u^3 + 6 u^2 + 4u + 1$.

It takes about seconds to do using the binomial theorem. A bit longer of course if you expand as $\displaystyle (u + 1)^2 (u + 1)^2 = (u^2 + 2u + 1)(u^2 + 2u + 1) = ......$

So the integrand is $\displaystyle u + 4 + \frac{6}{u} + \frac{4}{u^2} + \frac{1}{u^3}$.

The integral is therefore $\displaystyle \frac{u^2}{2} + 4u + 6 \ln |u| - \frac{4}{u} - \frac{1}{2 u^2} + C$.

Now sub back u = x - 1. Done. (Although $\displaystyle \frac{(x - 1)^2}{2} + 4(x - 1)$ can potentially be expanded and simplified).

We're all fortunate that passionate experts like Krizalid are available to help - for free (except for the small price of putting up with a condescending tone )

7. I don't think Krizalid's comment was condescending at all. I think it was good advice.

-Dan

8. My apologies. As was hinted at, I was a little irritated at this problem because when I get an answer that's really length and ugly, I tend to question whether it's right, and unfortunately a lot of the homework problems my teacher assigns are not in the solutions manual so I have no method to check them outside here.