Question

integration of logx dx

Answer 1

$\int{1.logx}.dx$
this can be solved easily with help of 'integration by part ' .

by ILATE concept,
logx is the first Function
1 is the 2nd function

now, use the concept
$I \: = \int{f_1.f_2}dx=f_1.\int{f_2}dx-\int{(\int{f_2}}dx).(\frac{df_1}{dx})dx$
$then, I = \int{1.logx}dx =logx\int{1}dx-\int(\int{1}dx).(\frac{logx}{dx})dx$
I = x.logx - x + C