Check my Integration Homework: 1/x dx = (1/3)ln(ln(x)) + c

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Homework Statement



http://i.minus.com/j5qAhbyrxc1jX.jpg

Homework Equations



∫1/x dx = ln(x)

The Attempt at a Solution



Can you guys check my work? I got the solution above and I've been pounding my head against the wall for the last two hours as to why that solution, according to my book, is incorrect.

The solution, according to the book, is

(1/3)ln(ln(x)) + c
 
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If you use that ln(x^3)=3*ln(x) you can show that your answer is the same as the books, with an extra additive constant.
 
Both answers are correct. If you differentiate both answers, you wind up with the integrand.


Tip: Your work would have been simpler if you had written ln(x3) as 3ln(x).
 
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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