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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 $\frac{d}{dx} \log x = \frac{1}{x}$. – Pragabhava Nov 26 '12 at 17:30
Yes, I figured that ou, still very tough in my eyes.. – Jijbentlekker Nov 26 '12 at 17:32
up vote 3 down vote accepted

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

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.. – Jijbentlekker Nov 26 '12 at 18:14
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 $d/dx(ln(x)) = 1/x$). – amWhy Nov 26 '12 at 18:20
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? – Jijbentlekker Nov 26 '12 at 18:30
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. – amWhy Nov 26 '12 at 18:35
The du clarifies the variable with respect to which we are integrating. – amWhy Nov 26 '12 at 18:35

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


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

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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). – amWhy Nov 26 '12 at 18:54
Oh I see, thank you, I fully understand it now. – Jijbentlekker Nov 26 '12 at 19:02
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: – amWhy Nov 26 '12 at 19:06

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? – Jijbentlekker Nov 26 '12 at 18:16
I don't think I'd call it "arithmetic", but otherwise that is correct. – Michael Hardy Nov 27 '12 at 0:36

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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