# Integral of $\frac{\ln (x^2)}{x}$

I'm not sure where to start with this problem. I have the answer, but do not know how to work it out. I always get confused when it comes to integrating natural logs. Do I use u substitution or integration by parts? I've tried working it out with both, but I'm not sure what to use as u. I tried $(\ln x)^2$ as u, but ended up with $\frac{(\ln x)^3}{3} + C$ when it should be $(\ln x)^2 + C$ (I apologize for improper format).

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Do you mean $\ln(x^2)$ or $(\ln x)^2$? –  Michael Albanese Jan 6 '13 at 6:47
The question is unclear, but I assume it is (ln x)^2 –  Jaden M. Jan 6 '13 at 6:50
@JadenM.: Your title seems to have a different meaning. I don't think that assumption is safe. –  Jonas Meyer Jan 6 '13 at 7:05
I simply do not understand which one it is. It wasn't specified in the problem, so I have no idea whether to use (lnx)^2 or ln(x)^2. –  Jaden M. Jan 6 '13 at 7:28
@CBenni: If by "he" you mean Jaden M., then I disagree, because Jaden M. did not know what it meant, and did not know how the answer was obtained. If you meant what is intended in the problem, then I agree, as mentioned in my answer, and as Amr mentioned in a comment. –  Jonas Meyer Jan 6 '13 at 20:38

Note that $\frac{1}{x}=\frac{d}{dx}\log(x)$, therefore: $$\int \frac{(\log (x))^2}{x}dx=\frac{(\log(x))^3}{3}+C$$ In case you meant $\log(x^2)$, check Babak's answer.

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When I did the problem, that is the answer I got as well. However, the answer should be (lnx)^2 + C. I don't understand how. –  Jaden M. Jan 6 '13 at 6:54
Then you wanted to integrate $\log(x^2)$ and not $(\log(x))^2$ –  Amr Jan 6 '13 at 6:55
@Jaden M.: If you want to see whether the answer is correct, you can differentiate it. The derivative of $\frac{1}{3}(\ln(x))^3$ is $\dfrac{(\ln(x))^2}{x}$, and the derivative of $(\ln(x))^2$ is $\dfrac{2\ln(x)}{x}$. –  Jonas Meyer Jan 6 '13 at 7:01

If you have $\int\frac{\ln(x^2)}{x}dx$ then note that $\ln(a^b)=b\ln(a), a>0$ so you have $\int\frac{\ln(x^2)}{x}=\int\frac{2\ln(x)}{x}dx$ and this latter one can be evaluated by setting $\ln(x)=t$. For another one look at @Amr's answer.

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Well done, again!+1 –  amWhy Feb 23 '13 at 0:05

If you are trying to find an explicit formula for the antiderivative of a function given by an explicit formula, it is essential that you know what the explicitly given formula means. While you could solve the problem both ways for additional practice, clearing up what the actual problem was in the first place is still needed.

When notation is ambiguous, there is not generally a correct mathematical way to figure out what the notation means. It requires some way of finding out what conventions are used in the source of the problem. If the problem is in a textbook, you often find explanations of the conventions earlier in the book, or at least enough context from earlier examples to determine the conventions. If the problem came from a teacher, you can ask the teacher what is meant. If the problem arises in the real world, you can look at the context of where it came from, and you may want to kindly ask the person who wrote it down to use less ambiguous notation in the future.

In calculus textbooks, it is common to denote powers of some functions by putting the exponent between the function name and the argument. This allows less parentheses to be written, and thus laziness may be a key source of the popularity of the convention. This is most commonly seen for trigonometric functions. E.g. $\sin^4 x$ means $(\sin(x))^4$. When this convention is used, appearance of something like $\sin x^4$ can usually be safely inferred to mean $\sin(x^4)$. The reason it would be expected to not mean $(\sin(x))^4$ is that the latter would be written instead as $\sin^4 x$.

The same convention may be applied to logarithms, and thus $\ln^2 x$ could mean $(\ln(x))^2$ while $\ln x^2$ would mean $\ln(x^2)$, which turns out to be equivalent to $2\ln(|x|)$ for all real nonzero $x$. The context of what you say the answer "should be" indicates that $\ln x^2=\ln(x^2)$ is likely to be the convention used in your problem, whereas $(\ln x)^2$ is used when the function is squared rather than the argument. (So while I don't see evidence of your source using the "$\ln^2 x$" notation, they do seem to keep the convention that exponents after the argument without parentheses apply to the argument, not the function.)

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Can be solved using substitution method:
Edit: $ln{(x^2)}$ can be written as $2ln{x}$. (Basic properties of logarithms)
So the main equation now looks like: $\int\frac{2ln{x}}{x}dx$
Now, the substitution part:
Put $ln{x} = t$
differntiating both the sides:
$\tfrac{dln{x}}{dx} = \tfrac{dt}{dx}$
=> $\frac{1}{x}dx = 1dt$
Now substitute $ln{x} = t$ and $\frac{dx}{x} = dt$ in the main expression
The expression so formed is: $\int 2tdt$
Now, it's easier to integrate this.
We get: $2\frac{t^{2}}{2}$ + C = $t^{2}$ + C
Now substitute $ln{x}$ back with $t$
We get $(ln{x})^2$ + C.

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How I would interpret "$\ln(x)^2$", it cannot be written as $2\ln x$. Unless you meant $\ln(x^2)$ (and $x>0$), but $\ln(x)^2$ would reasonably be interpreted to mean $(\ln x)^2$. That is, those parentheses serve the purpose of showing what is the input of the logarithm. A major problem the OP had was interpreting what was even meant by the notation, so using $\ln(x)^2$ to mean $\ln(x^2)$ might not help. Otherwise, I agree with this as a solution to what is probably the intended problem, and it gives more detail on the outline of Babak's answer. (To me, $\ln(x)^2=(\ln x)^2\neq \ln(x^2)$.) –  Jonas Meyer Jan 16 '13 at 20:54
@JonasMeyer Ooops... Sorry. I mis-typed it. Since, in the question, it is $ln({x}^2)$ –  xyres Jan 17 '13 at 3:21
Thanks. "In the question, it is $\ln(x^2)$." That depends on what part of the question you are looking at. –  Jonas Meyer Jan 17 '13 at 3:26
@JonasMeyer Oh! The question is in the "title": $\frac{ln(x^2)}{x}$. And the desired answer is $(ln{x})^2$. –  xyres Jan 17 '13 at 3:30