Question

Evaluate the integral or show that it is divergent \[\int\limits_{1}^{e} dx / x \sqrt{lnx}\]

Answer 1

http://www.wolframalpha.com/input/?i=integrate+x+from+1+to+E+x+Sqrt%5BLog%5Bx%5D%5D

Answer 2

ok so me know its improper
\[\int\limits_{}^{}\frac{dx}{x \sqrt{lnx}},u=lnx=>du=\frac{1}{x} dx\]
\[\int\limits_{}^{}\frac{1}{\sqrt{u}} du=\int\limits_{}^{}u^\frac{-1}{2} du\]

Answer 3

\[=\frac{u^{\frac{-1}{2}+1}}{\frac{-1}{2}+1} +C\]

Answer 4

so so lets get back to the improper integral

Answer 5

2u^{1/2}

Answer 6

you are right

Answer 7

forgot to flip

Answer 8

\[2u^\frac{1}{2}+C=2(lnx)^\frac{1}{2}+C\]
\[\lim_{n \rightarrow 1^+} (2(lnx)^\frac{1}{2})|^e_n\]

Answer 9

\[2\lim_{n \rightarrow 1^+}[(\ln(e))^\frac{1}{2}-(\ln(n))^\frac{1}{2}]\]

Answer 10

\[2-2\lim_{n \rightarrow 1^+ }(\sqrt{lnx})\]

Answer 11

we can use direct substitution :)

Answer 12

=2

Answer 13

yep

Answer 14

its getting late

Answer 15

yep...bed time for me

Answer 16

me too
goodnight
like doing this when i have a partner lol

Answer 17

or more i like doing it more when i have a partner

Answer 18

yep

Answer 19

night

Answer 20

peace