Help with an integration problem involving the natural log 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? 
 A: $$\int \dfrac{(\ln(x))^2}{x} dx \;=  \;\int (\ln(x))^2 \cdot \dfrac 1x dx $$
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 $x$:$$\dfrac{(\ln(x))^3}{3} + C$$
A: 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?
A: Hint: directly
$$\int f(x)^nf'(x)dx=\frac{f(x)^{n+1}}{n+1}+K$$
A: 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:
$$
\int u^2 \, du.
$$
A: 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 $\int u^2 du = \dfrac{1}{3}u^3 + c = \dfrac{1}{3}\ln (x)^3 + c$
