What is #int ln(x)/x#?

1 Answer
Nov 30, 2015

#(lnx)^2/(2) +C#

Explanation:

#intln(x)/x dx hArr int[ln(x)*(1/x)]dx #

Can be solve by u-substitution

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

#" " " " du = 1/x dx " " " " (2)" " #

equal to #dx# in the original problem

Let substitute that back into the original
#int u du = (u^2)/2 +C " " " " " (3)" #

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

Final answer: #(lnx)^2/(2) +C#

I hope this help :)