Integral of $1/x^2$ $$\int\frac{1}{x^2}dx$$
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 $$\frac{ln(x^2)}{2x}+ C$$
 A: "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!
A: 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.
A: 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 $$\frac{f'(x)}{f(x)}=\frac{e^{\frac{-1}{x}}\frac{1}{x^2}}{e^{\frac{-1}{x}}}=\frac{1}{x^2}.$$
