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So I'm currently trying to solve

$$\int \sqrt{ 1+\frac{1}{3x} } \, \, dx$$

I know that this can also be represented as ((x+1/3)/x)^1/2 but I dont like that form. I also know that this can be done with sustitution. I've done lot's of stuff but I get stuck everytime. You don't need to solve me the problem, just point me to the right direction if you want to and I'll finish it myself.

I'll write my first impression. I proceed to choose $u = x + \frac{1}{3x^{2}}$

so $du =1 -1/3x^{2} dx$

So I end with $\int \sqrt{u} \, \, \, du -3x^{2} $

I'm sure this is wrong but I don't know why. Maybe it wasn't wise to choose u as the whole square root? I did so because the immediate integral of x^1/2 is easy. Did I do something wrong? Or can I continue from here? If so, how?

Thanks a ton!!! =)

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    if you use that u, make sure you get the power right: $du = 1-\frac{1}{3}x^{-2} dx$2012-10-13
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    Both the OP and @Alex write the right hand side of $du$ without parentheses, and it is wrong in my opinion.2012-10-14
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    @enzoib, yes it is, sorry. can't edit anymore though2012-10-14

4 Answers 4

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$$\int \sqrt{ 1+\frac{1}{3x} } \, \, dx$$ let $u^2=3x \implies 2u\ du = 3\ dx$ $$\frac{2}{3}\int \sqrt{ 1+\frac{1}{u^2} } \, \, u\ du$$ $$\frac{2}{3}\int \sqrt{ u^2+1 } \, \, du$$ now you can use $u = \sinh(t)$

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    This works nicely, unless we are worried about negative $x$, in which case we need to do a cases analysis.2012-10-13
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    you're my effing hero2012-10-13
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I suggest you to set $u = 3x$, but prior to that, do some thing with the sum under the root. Then use a pretty elementary integration technique.

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    So I did, and I ended with 1/3 integrate ((u+1)/u)^1/2 du.... and I'm stuck again =(2012-10-13
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    Yep, now split this expression in some imaginative way...2012-10-13
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    okay I'm trying, I'll let you know. Thanks a lot!! =)2012-10-13
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    Look at André Nicolás' idea too, it seems to me that using parts integration after his final statement would work.2012-10-13
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    I forgot to refresh the page lol =P. It's way over my level though. Anyway the only way I can split the expression is by distributing the roots. I can't find another way (I'll keep thinking though).2012-10-13
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    This sounds correct to me, now recall that $\left( \sqrt{x} \right)^{\prime} = \frac{1}{2 \sqrt{x}}$. Maybe there are still some issues with this 1 that keeps hanging, but I guess it can be solved in some easy way (I haven't done the integral myself, just roughly calculating it on my head).2012-10-13
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    I've no idea why is that useful. I'll keep it on my head though and I'll keep trying. Thanks a lot busman! I really appreciate it2012-10-13
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    Well, it is useful because you have a function times its derivative, so the integral becomes immediate. Anyway, you're welcome and I feel happy to help.2012-10-13
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As suggested by busman, it may be a good idea to let $u=3x$. The constants one has to drag around are then less annoying. So, apart from a constant factor, we want $$\int\sqrt{1+\frac{1}{u}}\,du.$$ Now I would suggest letting $w^2=1+\dfrac{1}{u}$. Then $$2w\,dw=-\frac{du}{u^2}=-(w^2-1)^2 \,du.$$ We end up having to find something like $$\int-\frac{2w^2\,dw}{(w^2-1)^2}.$$ This is a not completely pleasant partial fractions problem.

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    WOW man, that's awesome!! That's waaay over my level though =P2012-10-13
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    One can do better. As suggested by busman in a comment, one can alternately integrate by parts, one part $-2w$, the other part $\frac{w}{(w^2-1)^2}$. Then we end up basically wanting $\int \frac{w^2\,dw}{w^2-1}$, which by division basically leaves us with $\int\frac{dw}{w^2-1}$, easy partial fractions. Instead of my $w^2$, we could make the hyperbolic substitution $\sinh^2 y=1+\frac{1}{u}$, works nicely but too magical.2012-10-13
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Well, the problem appears to be the radical. I'm not sure setting $u$ equal to everything under the radical looks promising, as you still need a $du$. There is, however, more than one way to get rid of a radical.

If you are comfortable with hyperbolic functions, letting $\frac1{3x}=\sinh^2 u,x=\frac13\operatorname{csch}^2u$. Myself, I was never taught those and would go with the similar trig substitution $\frac1{3x}=\tan^2u,x=\frac13\cot^2u,dx=-\frac23\cot u\csc^2udu$.