Possible Duplicate:
Weird and difficult integral: $\sqrt{1+\frac{1}{3x}} \, dx$
So a couple of weeks ago I ran by accident with this integral. This is by no means homework, I just want to solve it out of curiosity. After asking for help and struggling for like 10 hours total I finally managed to reach something that makes some sense. However, the result is not the same as wolframalpha (integration or differentiate the primitive), and I was wondering where did I go wrong. Any help will be very much appreciated!
My work was:
$\int \sqrt{1+\frac{1}{3x}} \, \, dx$
$u^2 = 3x$
$2u \: du = 3 \:dx \rightarrow \frac{2}{3} u \: du = dx$
$\frac{2}{3}\int \sqrt{1+\frac{1}{u^2}} \, \, u \: du$
$\frac{2}{3}\int \sqrt{u^2 + 1} \, \, \: du$
$ \tan(s) = u \rightarrow du = \sec^2(s) ds$
$\frac{2}{3}\int \sqrt{\tan^2(s) + 1} \, \, \: \sec^2(s) \: ds$
Given that
$\tan(s) + 1 = \sec^2(s)$
then
$\frac{2}{3}\int \sqrt{\sec^2(s)} \, \, \: \sec^2(s) \: ds$
$\frac{2}{3}\int\sec^3(s) \: ds$
Integrate
$\frac{2}{3} \frac{1}{2} (\tan(s) \sec(s) + \log(\tan(s)+\sec(s)))$
Given that
$\tan(s) + 1 = \sec^2(s)$
then
$\sec(s) = \sqrt{\tan(s) + 1}$
therefore
$\frac{1}{3} (\tan(s) \: \sqrt{\tan(s) + 1} + \log(\tan(s)+\sqrt{\tan(s) + 1}))$
unsubstitute
$\frac{1}{3} (u \: \sqrt{u + 1} + \log(u+\sqrt{u + 1}))$
remember
$ u = \sqrt{3x}$
so
$\frac{1}{3} (\sqrt{3x} \: \sqrt{\sqrt{3x} + 1} + \log(\sqrt{3x}+\sqrt{\sqrt{3x} + 1}))$
You could do some algebra but by now it's too different from wolfram alpha. So anyway.. where did I go wrong?
Thanks a ton everyone!! =)