How do you find the definite integral for: # dx / (x sqrt(lnx))# for the intervals #[1, e^8]#?

Question

2285 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-06T13:40:15

# 4sqrt2#

first thing is a bit of pattern matching. i'll do it long-hand, this should be a lot briefer in practise....

notice that by the power and chain rules:

#d/dx(sqrt(f(x)) ) = 1/2 1/sqrt(f(x)) f'(x)#

so #d/dx(sqrt(ln x )) = 1/2 1/sqrt(ln x) 1/x#

and so # d/dx(color(red)(2) sqrt(ln x) ) = 1/sqrt(ln x) 1/x#

which means that here we have

# int_1^(e^8) d/dx( 2sqrt(ln x )) dx#

#= 2 [ sqrt(ln x ) ]_1^(e^8) = 2sqrt8 = 4sqrt2#