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My book gave the following problem: $$\int\frac{\ln(x)}x \text{ dx} $$ It seems to be a simple u-substitution problem, but I had a slight difficulty:
I set $u=\ln(x)$. Then, I had $\int\frac{u}x \text{ dx} $ I took the derivative of ln($x$) which is $\frac1x$. Then, I got dx = $x$ du. If I plug this in, I still have an $x$ in my integral. Where did I go wrong?

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    $\begingroup$ No, if you plug $dx=xdu$ into your integral $\int\frac uxdx$ you'll get $\int udu$, so no $x$ left. $\endgroup$ Jul 24 '19 at 14:07
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    $\begingroup$ The $x$'s cancel each other out $\endgroup$
    – Burt
    Jul 24 '19 at 14:10
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    $\begingroup$ Another way of thinking about it is: $$u = \ln(x) \Longrightarrow du = \dfrac{dx}{x}$$ So, when substituting: $$\int \underbrace{\left( \ln(x)\right)}_{u} \underbrace{\left(\dfrac{dx}{x} \right) }_{du} = \int udu$$ $\endgroup$ Jul 24 '19 at 14:11
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You have $du = \frac{dx}x$, so the integral becomes $$\int \frac{\ln(x)}{x} \, dx = \int u \, du.$$

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If $u=\ln x$, then $e^u=x$. So $\frac{\ln x}{x}$ becomes $\frac u{e^u}$ and, since $x=e^u$, $\mathrm dx$ becomes $e^u\,\mathrm du$.

Don't forget: you have to do the substitution everywhere, not just at some places.

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Using $$t=\ln(x)$$ then $$dt=\frac{1}{x}dx$$ then we get $$\int tdt=\frac{t^2}{2}+C$$

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