What is $\int \frac{ln (x)}{x}$ $int ln(x)/x$ ?

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Answer 1 · 2015-11-30T01:08:14

$\frac{{(ln x)}^{2}}{2} + C$ $(lnx)^2/(2) +C$

$\int \frac{ln (x)}{x} d x \Leftrightarrow \int [ln (x) \cdot (\frac{1}{x})] d x$ $intln(x)/x dx hArr int[ln(x)*(1/x)]dx$

Can be solve by u-substitution

Let $u = ln (x) (1)$ $" " u= ln (x) " " " " " (1) " "$

$d u = \frac{1}{x} d x (2)$ $" " " " du = 1/x dx " " " " (2)" "$

equal to $d x$ $dx$ in the original problem

Let substitute that back into the original
$\int u d u = \frac{u^{2}}{2} + C (3)$ $int u du = (u^2)/2 +C " " " " " (3)"$

Re-substitute $u = ln x$ $u= ln x$ into solution above (3)

Final answer: $\frac{{(ln x)}^{2}}{2} + C$ $(lnx)^2/(2) +C$

I hope this help :)

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