Why can't I use u-sub for this Particular Integral?

Question

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

$$\int\frac{8}{x(x+2)^3}\;dx$$

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!

Answer 1

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.