Why can't you apply the natural logarithm rule to integrate $\int \frac{1}{\sqrt{x}}dx$? I understand that $\frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}$, or $\frac{1}{x^2} = x^{-2}$, but why wouldn't you be able anyhow to apply the rule for which $\int \frac{1}{x}dx = \ln{|x|} + C$, and have, for example $\int \frac{1}{x^2}dx = \ln{|x^2|} + C$?
 A: Notice:


*

*$$\frac{1}{\sqrt{x}}=\frac{1}{x^{\frac{1}{2}}}=x^{-\frac{1}{2}}$$


So:
$$\int\frac{1}{x^n}\space\text{d}x=\int x^{-n}\space\text{d}x=\frac{x^{1-n}}{1-n}+\text{C}$$
Set $n=\frac{1}{2}$:
$$\int\frac{1}{\sqrt{x}}\space\text{d}x=\int\frac{1}{x^{\frac{1}{2}}}\space\text{d}x=\int x^{-\frac{1}{2}}\space\text{d}x=\frac{x^{1-\left(\frac{1}{2}\right)}}{1-\left(\frac{1}{2}\right)}+\text{C}=2\sqrt{x}+\text{C}$$

Now if we use the same rule with $n=1$, we get:
$$\int\frac{1}{x^1}\space\text{d}x=\int x^{-1}\space\text{d}x=\frac{x^{1-1}}{1-1}+\text{C}=\frac{x^{0}}{0}+\text{C}$$
$$\color{red}{\text{And we can't divide by}\space0}$$
And if we know $\frac{\text{d}}{\text{d}x}\ln(x)=\frac{1}{x}$:
$$\int\frac{\text{d}}{\text{d}x}\ln(x)\space\text{d}x=\int\frac{1}{x}\space\text{d}x$$
A: If I well understand you ask why, if we have $F'(x)=f(x)$ than we cannot have:
$$
\int g(f(x)) dx = g(F(x))
$$
You can see that this is impossible by using the chain rule:
$$
\frac{d}{dx}g(F(x))=g'(x)f(x) \ne  g(f(x))
$$
In the same way you can show that
$$
\int f(g(x))dx= F(g(x)) 
$$
does not work.
A: Because that ignores the chain rule.
$$
\frac d {dx} \ln(x^5) = \frac 1 {x^5}\cdot \frac d {dx} x^5 = \frac 1 {x^5} \cdot 5x^4 = \frac 5 x \ne \frac 1 {x^5}.
$$
Notice that one can also do this:
$$
\frac d {dx} \ln (x^5) = \frac d {dx} (5\ln x) = 5\frac d {dx} \ln x = 5\cdot \frac 1 x = \frac 5 x.
$$
If the proposed derivative $\dfrac 1 {x^5}$ were right, then you'd have to wonder why these two methods don't yield identical results.
A: When integrating any function $f(x)$ you must take into account the variable along which you are integrating, in this case $d x$. 
So, according to your rule we can also state that 
\begin{equation}
\int \cos x^2 d x = \sin x^2 + C
\end{equation} 
, since $\displaystyle \int \cos x d x = \sin x + C$
As long as it is incorrect, you should solve these types of integral using various methods such as change of variable to have the "correct" result.
A: Mainly because x and x^2 are not the same thing. It'll be much better if you rewrite f(x) as x^-2, which will be easier to compute. 
