What is the integral of $ln (\sqrt{x})$ $ln(sqrt(x))$ ?

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Answer 1 · 2015-05-02T18:06:00

By part :

$\int ln (\sqrt{x}) d x$ $intln(sqrt(x))dx$

$d u = 1$ $du = 1$
$u = x$ $u = x$

$v = ln (\sqrt{x})$ $v = ln(sqrt(x))$
$d v = \frac{1}{2 x}$ $dv = 1/(2x)$

$[x ln (\sqrt{x})] - \frac{1}{2} \int d x$ $[xln(sqrt(x))]-1/2intdx$

$[x ln (\sqrt{x}) - \frac{1}{2} x]$ $[xln(sqrt(x))-1/2x]$

don't forget $ln (a^{b}) = b ln (a)$ $ln(a^b) = bln(a)$

$[\frac{1}{2} x ln (x) - \frac{1}{2} x]$ $[1/2xln(x)-1/2x]$

factorize by $\frac{1}{2} x$ $1/2x$ and don't forget the constant !

$[\frac{1}{2} x (ln (x) - 1) + C]$ $[1/2x(ln(x)-1)+C]$

What is the integral of ln(√x)ln(sqrt(x))?