Basic Integration Solved: Find Antiderivative of ln(x^20)^2

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[SOLVED] Basic Integration

\int (ln(x^{20}))^{2}dx

Just need to find an antiderivative.

Once again, drawing a blank on the easy stuff. Could anybody tell me which method of integration to use?
 
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Hrmmm..After playing with it for a long time... I feel like a middle schooler.

Partial integration didn't help. Couldn't find a working substitution. No trigonometric etc

So that means (this is just a guess) that I'm probably missing a simple little step at the beginning that is throwing me off. Been searching through little identities and looking at examples in my textbook etc.

No luck =(

Any help would be great!
 
Um... think about the law of the log function.

ln(x^2) = ?
 
Ughh...I knew it. Forgot to change ln(x^20) into 20lnx.

All that time wasted...lol

Thanks tho!
 
Lanza52 said:
Ughh...I knew it. Forgot to change ln(x^20) into 20lnx.

All that time wasted...lol

Thanks tho!

You would have wasted more time if you didn't come here!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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