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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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    @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 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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    @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