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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!! =)

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    I changed $sec$, $tan$, and $log$ to $\sec$, $\tan$, and $\log$. This not only makes them non-italicized but also provides proper spacing in expressions like $x\sin x$.2012-11-02

1 Answers 1

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In your first change of variable $u^2=3x$, $u^2$ is non-negative but in the original integral $x$ can take negative values.

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    +1! Yes, I know that but why is the result so different from wolframalpha then? =(2012-11-02