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I have to find this integral $$\int\frac{2\ln(x)}{x}dx$$ This is how I began: $$\int\frac{2\ln(x)}{x}dx=2\int\frac{\ln(x)}{x}dx$$ Then I tried substitution $e^u=x$ to get $u=\ln(x)\longrightarrow du=\frac1{dx}\rightarrow dx=\frac1{du}$

$$2\int\frac{\ln(x)}{x}dx=2\int\frac{\ln(e^u)}{e^u}\frac1{dx}$$ But then it gets complicated, and I'm not sure if the steps I've done are right. Need some help guys.

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Correction: $du = \frac{1}{x}dx$ and use $ln(e^u) = u$ – tpb261 Jun 16 '14 at 11:25

4 Answers 4

up vote 2 down vote accepted

$$\ln x=u\Rightarrow \frac{1}{x}dx=du\Rightarrow\int\frac{2\ln(x)}{x}dx=2\int udu=u^2+c=\ln^2x+c$$

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but why ln(x)=u => dx/x=du and not 1/dx=du? – user157374 Jun 16 '14 at 11:26
@user157374 Because $\frac{du}{dx} = \frac{1}{x}$ – 5xum Jun 16 '14 at 11:27
Ah right, get it now! thanks :) – user157374 Jun 16 '14 at 11:28

$$\ln x=y\implies \frac{dx}x=dy\implies\int\frac{\ln x}x\ dx=\int y\ dy$$

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It should be $du = \frac1x dx $

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Why?? Derivative ln(x) is 1/x – user157374 Jun 16 '14 at 11:21

Hint :

Rewrite the integral as $$ 2\int\ln x\ d(\ln x). $$

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How please? I can't seem to recognize that – user157374 Jun 16 '14 at 11:23
@user157374 $\dfrac{d}{dx}(\ln x)=\dfrac1x\quad\Rightarrow\quad d(\ln x)=\dfrac1x\ dx$. – Tunk-Fey Jun 16 '14 at 11:25

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