As I continue studying for my upcoming exam, I stumbled across an Integral that I found quite interesting. It is as follows:


Now, my logical self tells me to go straight for partial fraction decomp, but I figured hey, why not see if I can u-sub this? It seemed like I could, with $u=x+2$, $x=u-2$ and $du=dx$. Pulling the constant out, I have:

$$8\int\frac{du}{(u-2)u^3}$$ Afte distributing, I got $$8\int\frac{du}{u^4-2u^3}$$

After integrating, I was left with: $$-\frac{8}{3(x+2)^3}+\frac{8}{(x+2)^2}\;+C$$

Maybe I made a mistake somewhere, I'm not sure. I then went back and did the problem using my original idea(partial fraction decomp) and yielded the correct answer of $$\ln|x|-\ln|x+2|+\frac{2}{(x+2)}+\frac{4}{(x+2)^2}\;+C$$

These answers don't look anything alike, and when I took the derivative of my original answer and it definitely did not match.

So my question is, can I use u-sub in this instance as a valid technique, and if so, where did I go wrong?

If this is not a suitable situation to use u-sub, please help me understand why because it looked like it would work out so nicely. I just want to understand the process behind everything as much as I can. Thanks!

  • $\begingroup$ Write $$2^3=(x+2-x)^3$$ $\endgroup$ – lab bhattacharjee Jun 16 '16 at 8:21
  • $\begingroup$ you can't break up the denominator in this way. you have attempted to put $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$ $\endgroup$ – David Quinn Jun 16 '16 at 8:23
  • 1
    $\begingroup$ Shoot, you are so right. If I had a dollar for every time I've made that mistake... $\endgroup$ – FuegoJohnson Jun 16 '16 at 8:24

You can use a $u$-subsitution, but you'll be left with a very similar integral that will still require a partial fraction decomposition. So there's nothing to gain really, from first performing that substitution.

Your error lies here:

$$8\int\frac{du}{(u-2)u^3}$$ After distributing, I got $$8\int\frac{du}{u^4-2u^3}$$ After integrating, I was left with: $$\color{red}{-\frac{8}{3(x+2)^3}+\frac{8}{(x+2)^2}\;+C}$$

Distributing is allowed, but doesn't help: you'll want to do a partial fraction decomposition on the initial integral in the variable $u$ as well.

  • $\begingroup$ Yes, @DavidQuinn pointed that out above but thank you for clarifying further. It's a mistake I make more than I should, despite the fact I am well aware of it (I just had this discussion w my professor last week about how i've been trying to improve that part of my algebra). Hopefully this is the final learning experience. Thanks for clarifying the downfall of that method as well, i'll avoid it in the future. $\endgroup$ – FuegoJohnson Jun 16 '16 at 8:29
  • $\begingroup$ I was already typing an answer but it's hard to 'beat' a fast comment in speed. The good news is you seem to be on track with the integration methods; the error was indeed purely algebra - but that of course, is not very good news ;-). $\endgroup$ – StackTD Jun 16 '16 at 8:32
  • $\begingroup$ Well I give you kudos, your formatting speed is impeccable ;) I'm getting faster. but it still takes me several mins to format a post. The whole $\frac{1}{a+b}=\frac1a + \frac1b$ thing has been my achilles heel a few times, but overall I try to be fairly meticulous with my algebra. Trig, though, is a different story. On a side note, how do you color your txt? Mathjax basic tutorial doesn't seem to have it... $\endgroup$ – FuegoJohnson Jun 16 '16 at 8:42
  • $\begingroup$ In math-mode, you use \color{red}{\sin x} to produce $\color{red}{\sin x}$. Other basic colors are available by name. $\endgroup$ – StackTD Jun 16 '16 at 8:43

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