# Is it legal to integrate $\frac{4 }{\sqrt{ x }}$ as $4\ln \sqrt x + C$ instead of $8\sqrt{x} + C$? Why / Why not?

I have the following integral:

$$\frac{4}{\sqrt x }$$

As far as I'm aware, this is equal to $4\frac{1}{\sqrt x}$.

Thus, can we solve it using the $\ln$ rule so the answer is $4 \ln \sqrt x$ ?

I know you can solve it without using the $\ln$ rule so the answer is $8 \sqrt x$.

However, by quickly checking myself, $4 \ln\sqrt x \ne 8\sqrt x$. If this is true, I assume I can't use the $\ln$ rule for this equation, so the question becomes: when can I use the $\ln$ rule where $\frac 1x$ = $\ln x$?

• The derivative of $\ln(\sqrt{x})$ is not equal to $1/\sqrt{x}$. You need to apply the chain rule and therefore multiply by the derivative of $\sqrt{x}$. – Oliv Mar 9 '17 at 10:16
• $1/\sqrt{x} = 1 / x ^ \frac{1}{2} = x^{-\frac{1}{2}}$ now use the powers rule, not the log rule (which would be dead wrong here) – Cato Mar 9 '17 at 10:24
• @user35508 please don't change the meaning of the post you're reformatting. Original didn't have + c in the body where you added it. – Ruslan Mar 9 '17 at 10:28
• @Ruslan.... Adding +C doesn't change meaning at all...Moreover, I think it improves the meaning as the OP is definitely evaluating an indefinite integral – user35508 Mar 9 '17 at 10:32
• @user35508 On the other hand, I think it's fine (and potentially very helpful) to point out errors like the missing $+C$ in a comment. – David K Mar 9 '17 at 14:32

You have learned that $$\int\frac{1}{x}\,dx=\ln x.$$ Yo seem to believe that for any function $f$ $$\int\frac{1}{f(x)}\,dx=\ln(f(x)).$$ Well, it is not true. You can check it by differentiating: $$(\ln(f(x)))'=\frac{f'(x)}{f(x)}\ne\frac{1}{f(x)}.$$ Let's see another extreme example. What is the integral of the constant function $f(x)=\dfrac12$? Is it $$\frac12\,x\quad\text{or}\quad \ln2?$$

It's wrong, from the fundamental theorem of calculus that is $f(x) = \dfrac{d}{dx}\int_0^xf(t) dt$ If you differentiate $F(x) = 4\ln{\sqrt{x}}$ you get $\dfrac{2}{x}\neq\dfrac{4}{\sqrt{x}}$.You can only use the $\ln{x}$ rule iff $f(x) =\dfrac{1}{x}$

No....You are doing it wrong

$$\frac{d\ln{\sqrt{x}}}{dx} \neq \frac{1}{\sqrt{x}}$$

You can verify this by checking with the Chain rule...

The original answer to your indefinite integral is in fact $8\sqrt{x}+C$

NOTE:- You can use $\ln(x)$ if the integrand is $\frac{1}{x}$ ..For your original question $\frac{1}{\sqrt{x}}=x^{-1/2}$ ..You can use the power rule i.e. $$\int x^n dx=\frac{x^{n+1}}{n+1}+C$$

• This does not answer the question. – Improve Mar 9 '17 at 10:24
• it doesn't answer the question, it is also a repeat of the first comment, it should have been a comment if anything – Cato Mar 9 '17 at 10:26
• Edited the answer – user35508 Mar 9 '17 at 10:31

First of all, please do not ever write things like "$\frac1x = \ln x.$" If you mean to say that the integral of $\frac1x$ is $\ln x,$ write "The integral of $\frac1x$ is $\ln x.$"

Second, you can use the $\ln$ rule when you are able to put the integral into exactly the format in which the $\ln$ rule was given. The same can be said for any integration rule. For example, if you have learned a rule is given in the form $$\int \frac1x \,dx = \ln x + C, \tag1$$ then you can always use it when integrating $\frac1x$ with respect to $x.$ You can also use it to integrate $\frac1y$ with respect to $y$: $$\int \frac1y \,dy = \ln y + C, \tag2$$ because we can get Equation $2$ from Equation $1$ by performing a U-substitution with $x = y.$ More generally, it will work for anything of the form $$\int \frac1\square \,d\square = \ln \square + C$$ provided that you put the exact same thing in all three boxes. For example, $$\int \frac1{\sqrt x} \,d{\sqrt x} = \ln {\sqrt x} + C.$$ But that's not what we usually mean by "the integral of $\frac1{\sqrt x}.$" Here is what we usually mean when we say that: $$\int \frac1{\sqrt x} \,dx.$$ Sure, that looks a bit like $\int \frac1\square \,d\square,$ but it does not have the exact same thing in both boxes, so you cannot apply a rule that requires both boxes to be filled with the same thing.