how do I deal with the square power in this problem:
Integrate: [ln(x))^2]

w/ U-sub or IBP
Am I able to use U-Sub if u=ln(x) ?

Question

how do I deal with the square power in this problem:
Integrate: [ln(x))^2]

w/ U-sub or IBP
Am I able to use U-Sub if u=ln(x) ?

waynex · Answer

IBP is my first thought, since it's the analog of the product rule. If you use u=ln(x), what is du? It's 1/x dx, and 1/x is not in that integral. u substitution doesn't look promising to me.

anonymous · Answer

It's almost like this --
please look at the picture below

waynex · Answer

alexuspanait, that rule for logarithms applies to $$\ln(x^a) = a \cdot \ln(x).$$
Note the difference between the following: $$\ln(x^a)  \neq \ln(x)^a.$$

anonymous · Answer

@Waynex - you're right , I was wrong , sorry, my bad
I tried to help ...

waynex · Answer

There is no need to be sorry. I only mentioned it to help avoid any confusion. Don't let that deter you. Best

anonymous · Answer

Well if you know you log properties which is the same for natural logs, then

$$\ln a ^b = b \ln a$$

so you end up getting...

$$\int\limits_{?}^{?} \ln x ^2 = \int\limits_{?}^{?} 2 \ln x$$

anonymous · Answer

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