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**DBA** I need to integrate

$\displaystyle

\int_1^4 \frac{1}{x \sqrt[3]{lnx}}dx

$

The integral is impoper because x cannot be 1. Since 1 is at the left hand, I rewrite it to...

$\displaystyle

\lim t \rightarrow 1^+ \int_t^4 \frac{1}{x \sqrt[3]{lnx}}dx

$

Then I integrate and get...

Let u=lnx, du=1/x dx

$\displaystyle

\int_1^4 \frac{1}{x \sqrt[3]{lnx}}dx = \frac{3}{2} (lnx)^{2/3} +C

$

Then

$\displaystyle

\lim t \rightarrow 1^+ \int_t^4 \frac{1}{x \sqrt[3]{lnx}}dx

$

=

$\displaystyle

\lim t \rightarrow 1^+ [\frac{3}{2}*[(ln4)^{2/3} - (lnt)^{2/3}]]

$

Now I do not know how to find out what the answer is. Lnt would be Ln1+, so it gets closer and closer to ln1. Ln1 = 0. Does the term (lnt)^2/3 gets zero then?

And the limit goes to 3/2((ln4)^{2/3} - 0 ?

Thanks for any help.