please help me with this integral, I can't figure out how to solve it from the manual.
$$ 4\pi\int_1^e \frac{\ln(x)}{x} \, \mathrm{d}x $$
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1When you say (lnx)' do you mean $\frac{1}{x}$ or is that just a typo, and the integrand is $ln^2(x)$? – Grid Apr 04 '14 at 19:28
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that's not a typo, it's 1/x – Mihai101 Apr 04 '14 at 19:29
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4Then it's trivial, let $u = ln(x)$ and $du = \frac{dx}{x}$. – Grid Apr 04 '14 at 19:30
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You should learn to use math markup also in titles! – kjetil b halvorsen Apr 04 '14 at 19:59
2 Answers
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$$4\pi\int_1^e(\ln x)(\ln'x)\,dx=4\pi\int_1^e \frac12(\ln^2 x)'\,dx=2\pi(\ln^2 e-\ln^2 1)=2\pi$$
user2345215
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$4\pi \cdot \int_{1}^e lnx \cdot (lnx)' dx = 4\pi \int_{1}^e \dfrac{lnx}{x} dx = \dfrac{1}{2} \cdot 4\pi \cdot ((lnx)^2)|_{x = 1}^{e} = 2\pi$
DeepSea
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i don't get it, how do you integrate lnx/x ?? where did the 1/2 came from – Mihai101 Apr 04 '14 at 19:34
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He used a u-substitution to get the integral to be of the form $\int u du = \frac{1}{2} u^2$ by the power-rule for integration. – Grid Apr 04 '14 at 19:46