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I'm new to integral calculus, I started literally 15 minutes ago, and I need help with this question:

$$\int \dfrac{\ln(x)^2}{x} dx $$

My first step was:

$$\int \dfrac{1}{x}\ln(x)^2 dx $$

However, what to do next, how to solve this using the reverse chain rule?

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    **Hint:** Use the fact that $\frac{d}{dx} \log x = \frac{1}{x}$.2012-11-26
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    Yes, I figured that ou, still very tough in my eyes..2012-11-26

5 Answers 5

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$$\int \dfrac{(\ln(x))^2}{x} dx \;= \;\int (\ln(x))^2 \cdot \dfrac 1x dx $$

Let $u = \ln(x).\;$ So $u^2 = (\ln(x))^2$

Then $\dfrac{du}{dx}(\ln(x)) = \dfrac1x$, so we can replace $ \dfrac1x dx$ with $du$.

By substitution, $$\int (\ln(x))^2\cdot \dfrac1x dx \;=\; \int u^2 du$$ Evaluating the integral gives

$$\dfrac{u^3}{3} + C$$

Then replacing $u$ with $\ln(x)$ gives us the integral in terms of $x$:$$\dfrac{(\ln(x))^3}{3} + C$$

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    So basically, we can replace $\dfrac{1}{x} dx$ with $du$ because $du=dx . \dfrac{1}{x}$? So it all boils down to basic arithmetic? I'm having troubling grasping this, because I'm self-studying integral calculus, while I studied diff. calc. at school, and we never actually manipulated the formulas like this..2012-11-26
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    Yes, integration by substitution (reverse chain rule) requires arithemetic, algebra, when necessary, and also a firm grasp of derivatives, so you can recognize appropriate substitutions (in this case, that $d/dx(ln(x)) = 1/x$).2012-11-26
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    So, when $du$ is in an integral it has no meaning? Because I always see that $du$ gets kind of disregarded? Or is that something I should just assume right now and learn about later?2012-11-26
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    It's just like dx when integrating with respect to x. With substitution, we are evaluating an integral with respect to u (hence du), then we "back substitute" to get the integrated formula back in terms of x.2012-11-26
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    The du clarifies the variable with respect to which we are integrating.2012-11-26
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Taking $u = \log x$, then $du = \frac{dx}{x}$, hence $$ \int \frac{\log(x)^2}{x} dx = \int u^2 du $$

Can you finish it?

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Hint: directly

$$\int f(x)^nf'(x)dx=\frac{f(x)^{n+1}}{n+1}+K$$

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Here's a hint: $$ \int (\ln x)^2 \Big( \frac1x\,dx\Big). $$ To understand how to use the "reverse chain rule", also called integration by substitution, is to understand this kind of hint.

The next step is to go from the hint above to this: $$ \int u^2 \, du. $$

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    So basically, we can replace $\dfrac{1}{x} dx$ with $du$ because $du=dx . \dfrac{1}{x}$? So it all boils down to basic arithmetic?2012-11-26
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    I don't think I'd call it "arithmetic", but otherwise that is correct.2012-11-27
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This is an attempt at answering my own question:

$$\int \dfrac{\ln(x)^2}{x} dx $$

We can rewrite that as:

$$\int \dfrac{1}{x} . \ln(x)^2 dx $$

Let $u = \ln(x)$, $\dfrac{du}{dx} = \dfrac{1}{x}$

$\dfrac{1}{x}.dx=du$

We get $\int u^2 du = \dfrac{1}{3}u^3 + c = \dfrac{1}{3}\ln (x)^3 + c$

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    Your last line is problematic: replacing $\dfrac1x dx$ with du$: recall that you are *multiplying* $ln(x)^2$ by $\dfrac1x dx$ so you need to *multiply* $u^2$ and du. You do not add du to the integral. Also, when integrating an indefinite integral, you need to include a variable, like "C" to denote a constant (e.g., see my post).2012-11-26
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    Oh I see, thank you, I fully understand it now.2012-11-26
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    You might want to check out the calculus offering at the Khan Academy (free tutorials on everything you'd want to know about calculus, including integration: http://www.khanacademy.org/2012-11-26