Why can't you apply the natural logarithm rule to integrate $\int \frac{1}{\sqrt{x}}dx$?

Question

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 $\int \frac{1}{x^2}dx = \ln{|x^2|} + C$?

Answer 1

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 $n=\frac{1}{2}$:

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

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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 $\frac{\text{d}}{\text{d}x}\ln(x)=\frac{1}{x}$:

$$\int\frac{\text{d}}{\text{d}x}\ln(x)\space\text{d}x=\int\frac{1}{x}\space\text{d}x$$

Answer 2

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.

Answer 3

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.

Answer 4

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.