2
$\begingroup$

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?

  • 0
    Yes, I figured th$a$t ou, still very tough in my eyes..2012-11-26

5 Answers 5

3

$\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 $x$:$\dfrac{(\ln(x))^3}{3} + C$

  • 0
    The$du$clarifies the variable with respect to which we are integrating.2012-11-26
2

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?

1

Hint: directly

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

0

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. $

  • 0
    I don't think I'd call it "arithmetic", but otherwise that is correct.2012-11-27
0

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

  • 0
    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