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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 $\frac 1x$ = $\ln x$?

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    $\begingroup$ 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 $\sqrt{x}$. $\endgroup$ – Oliv Mar 9 '17 at 10:16
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    $\begingroup$ $1/\sqrt{x} = 1 / x ^ \frac{1}{2} = x^{-\frac{1}{2}}$ now use the powers rule, not the log rule (which would be dead wrong here) $\endgroup$ – Cato Mar 9 '17 at 10:24
  • $\begingroup$ @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. $\endgroup$ – Ruslan Mar 9 '17 at 10:28
  • $\begingroup$ @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 $\endgroup$ – user35508 Mar 9 '17 at 10:32
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    $\begingroup$ @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. $\endgroup$ – David K Mar 9 '17 at 14:32
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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 $$ \frac12\,x\quad\text{or}\quad \ln2? $$

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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 $f(x) =\dfrac{1}{x}$

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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. $$\int x^n dx=\frac{x^{n+1}}{n+1}+C$$

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

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