1/x - integration by parts

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sallycats

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This has probably been done before, nor am I attempting to prove that 1=0, but I was unable to find other threads pertaining to this question, so here it is! :)

Use integration by parts to evaluate ?(1/x)dx.

Let u = 1/x, so du = -1/x[sup:3m2282ln]2[/sup:3m2282ln]dx
Let dv=dx, so v=x

Now ?(1/x)dx = uv - ?vdu = (1/x)(x) - ? (x)(-1/x[sup:3m2282ln]2[/sup:3m2282ln])dx = 1 + ?(1/x)dx

If ?(1/x)dx= 1 + ?(1/x)dx, then it seems that 0=1 which is obviously false.

My question is, what went wrong? How can this be possible?

I've given this some thought, and the conclusion I reached almost immediately is that when I did the above work, I did not consider a constant. If I add the constant when integrating, I still reach 0 + C = 1, or C = 1, which still does not seem to be right.

If my understanding is correct, C does not represent a specific number, such as 1. Adding C just accounts for the part of a function that has the derivative = 0, or in other words, any number.

Furthermore, would it be logical to conclude that ?(1/x)dx= 1 + ?(1/x)dx = 2 + ?(1/x)dx = 3 + ?(1/x)dx... ?
I would just need to continue integrating the right side of the equation, which would result with the original integral ?(1/x)dx with an additional +1 each time I integrate. Do each of these additions represent the constant?

I also know that ?(1/x)dx = ln|x|+C. How does that fit in with all of this? I feel like I am confusing myself, and could use a bit of guidance. :?
 
\(\displaystyle dv \ = \ dx, \ \implies \ \int dv \ = \ \int dx, \ \implies \ v+C_0 \ = \ x+C_1, \ implies \ v \ = \ x+C_1-C_0\)

\(\displaystyle Let \ C_2 \ = \ C_1-C_0, \ then \ v \ = \ x +C_2\)

\(\displaystyle Note: \ For \ the \ sake \ of \ brevity, \ these \ steps \ are \ usually \ omitted, \but \ in \ particular \ cases,\)

\(\displaystyle \int \frac{1}{x}dx, \ is \ one, \ they \ must \ be \ applied.\)
 
Hello, everyone!

\(\displaystyle \text{A similar fallacy: }\;\int\sin x\cos x\,dx\)

. . \(\displaystyle \int \sin x (\cos x\,dx) \;=\;\tfrac{1}{2}\cos^2\!x\quad[1]\)

. . \(\displaystyle \int \cos x(\sin x\,dx) \;=\;-\tfrac{1}{2}\sin^2\!x\quad [2]\)


\(\displaystyle \text{Equate [1] and [2]: }\;\tfrac{1}{2}\cos^2\!x \;=\;-\tfrac{1}{2}\sin^2\!x\)

. . . . . . . . . .\(\displaystyle \sin^2\!x + \cos^2\!x \;=\;0\)

. . . . \(\displaystyle \text{Therefore: }\qquad\qquad 1 \;=\;0\)

 
\(\displaystyle The \ point \ to \ remember \ when \ dealing \ with \ the \ above \ is \ a \ definite \ integral \ is \ just \ a \ number,\)

\(\displaystyle \ whereas \ an \ indefinite \ integral \ is \ a \ family \ of \ functions.\)

\(\displaystyle For \ the \ majority \ of \ cases, \ \int f(x)dx \ \ne \ \int f(x)dx \ unless \ their \ constants \ are \ equal.\)

\(\displaystyle And \ since \ each \ integral \ posesses \ an \ infinite \ number \ of \ constants, \ this \ possibility \ is \ slight, \ to\)

\(\displaystyle \ say \ the \ least.\)

\(\displaystyle An \ example: \ D_x\bigg[\frac{x^{2}}{2}+5\bigg] \ = \ x \ and \ D_x\bigg[\frac{x^{2}}{2}+2\bigg] \ = \ x\)

\(\displaystyle However, \ (first \ one) \ \int xdx \ \ne \ (second \ one) \int xdx \ as \ their \ constants \ differ.\)
 
Thank you everyone, this has been very helpful.
BigGlenntheHeavy said:
\(\displaystyle The \ point \ to \ remember \ when \ dealing \ with \ the \ above \ is \ a \ definite \ integral \ is \ just \ a \ number,\)

\(\displaystyle \ whereas \ an \ indefinite \ integral \ is \ a \ family \ of \ functions.\)
Click to expand...
I did not think of it this way, but this is indeed very important!
 
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