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$$\int\frac{1}{x^2}dx$$

For solving this we use the rule $f^m.f'$ making $f^m = x^{-2}$ thus the result is $$-\frac{1}{x}+C$$

My question is this:

Can I use the rule $\frac{f'}{f}$? If not, why not?

I was thinking in something like $f=x^2$ and $f'= 2x$. So it would become $$\frac{1}{2x}\int\frac{2x}{x^2}dx$$ and the result would be $$\frac{ln(x^2)}{2x}+ C$$

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    You can't factor $1/(2x)$ outside of the integral sign...2012-12-30
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    What is the $f^mf'$ rule? Or the $\frac{f'}{f}$ rule? I haven't seen this before.2012-12-30
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    @Limitless: The first is the power rule for integration, the second the log rule. That is, the first integrates to $f^{m+1}+C$, the second to $\ln f+C$.2012-12-30
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    @BrianM.Scott In that case, wouldn't he need to take $f=e^{\frac{1}{x^2}}$ in order for the log rule to work?2012-12-30
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    @Limitless: No: $\int\frac{2x}{x^2}~dx=\ln x^2+C$. The problem, of course, is that one can’t pull the $2x$ through the integral sign.2012-12-30

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"So it would become $$\frac{1}{2x}\int\frac{2x}{x^2}dx$$ ". This part is wrong. It would become $$\int\frac{1}{2x}\frac{2x}{x^2}dx$$ The $\frac1{2x}$ is a term of the integrand. Now if you were to use the rule $\frac{f^{\prime}}{f}$ rule the result would be $\ln\left|f\right|$. We thus want $$\ln\left|f(x)\right|=\frac{-1}{x}$$ or if you prefer, $$f(x)=e^{-\frac1x}$$ Observe that $$\frac{f^{\prime}(x)}{f(x)}=\frac{\frac{1}{x^2}\cdot e^{-\frac 1x}}{e^{-\frac1x}}=\frac{1}{x^2} $$ Therefore you can write $$\int\frac{1}{x^2}dx=\int\frac{ e^{-\frac 1x}}{ e^{-\frac 1x}x^2}dx$$ and apply the rule $\frac{f^{\prime}}{f}$. See if you can arrive at the correct result!

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    Thanks (again). So, correct me if I'm wrong. Only constants can go out the integral?2012-12-30
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    @Favolas Only constants!2012-12-30
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    @Nameless: even you needed to know (and **USE**) the solution to the integral found conventionally, in order to find the needed function to solve a much more cumbersome manner of trying to solve the integral in a different form. That's sort of like begging the question.2012-12-30
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    @amWhy I don't disagree. Why the use of capital and bold letters?2012-12-30
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    Not wanting to provoke discussion, but I wanted to just say this: amWhy has a good point. Personally, I found the approach tiresome. Why solve $f'/f=\frac{1}{x^2}$ and apply the "log rule" when you can simply go straight from $\int \frac{1}{x^2}dx$ to $-x^{-1}$ via $$\int x^mdx=\frac{x^{m+1}}{m+1}?$$2012-12-30
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    @Limitless I am not saying it is a better method. In fact it is cumbersome, artificial and not intuitive. However, the OP asked if it can be done that way and thats what I did. I see no point in further discussion2012-12-30
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    Thanks for making your stance clear. :-)2012-12-30
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Are you trying to rewrite $$ \int \frac{1}{2x} \frac{2x}x^2 \,dx = \frac{1}{2x} \int \frac{2x}x^2 \,dx ?$$ That's not valid --- you can't move expressions involving $x$ out from within the integral without doing something to justify it.

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It would appear you've used the "log rule" incorrectly. While it is true that $$\int \frac{f'(x)}{f(x)}dx=\ln\left(f(x)\right)+C,$$ we do not have that $$\int \frac{f'(x)}{f(x)}dx=\int \frac{1}{x^2}dx$$ when $f(x)=x^2$.

In order to properly apply the "log rule", you need $f(x)=e^{-\frac{1}{x}}$ since $$\frac{f'(x)}{f(x)}=\frac{e^{\frac{-1}{x}}\frac{1}{x^2}}{e^{\frac{-1}{x}}}=\frac{1}{x^2}.$$

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    Are you sure $$\int \ln(f(x))dx=\frac{f'(x)}{f(x)}+C,$$ because I always remembered it as $$\int \frac{f'(x)}{f(x)}dx=\ln(f(x))+C,$$...2012-12-30
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    @Nameless You're right. My mistake. I misunderstood Brian's comment.2012-12-30
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    @Nameless I will correct it for sake of posterity.2012-12-30
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    You mean prosperity?2012-12-30
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    @Nameless No. My statement meant that I was correcting it so that I would not confuse people who looked at it in the future. Prosperity is a consequence of correcting it, although.2012-12-30