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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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  • $\begingroup$ if you use that u, make sure you get the power right: $du = 1-\frac{1}{3}x^{-2} dx$ $\endgroup$
    – Alex
    Oct 13, 2012 at 18:32
  • $\begingroup$ Both the OP and @Alex write the right hand side of $du$ without parentheses, and it is wrong in my opinion. $\endgroup$
    – enzotib
    Oct 14, 2012 at 12:35
  • $\begingroup$ @enzoib, yes it is, sorry. can't edit anymore though $\endgroup$
    – Alex
    Oct 14, 2012 at 12:39

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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  • $\begingroup$ This works nicely, unless we are worried about negative $x$, in which case we need to do a cases analysis. $\endgroup$
    – André Nicolas
    Oct 13, 2012 at 18:56
  • $\begingroup$ you're my effing hero $\endgroup$
    – Gaspa79
    Oct 13, 2012 at 18:57
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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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  • $\begingroup$ So I did, and I ended with 1/3 integrate ((u+1)/u)^1/2 du.... and I'm stuck again =( $\endgroup$
    – Gaspa79
    Oct 13, 2012 at 18:36
  • $\begingroup$ Yep, now split this expression in some imaginative way... $\endgroup$
    – busman
    Oct 13, 2012 at 18:37
  • $\begingroup$ okay I'm trying, I'll let you know. Thanks a lot!! =) $\endgroup$
    – Gaspa79
    Oct 13, 2012 at 18:39
  • $\begingroup$ Look at André Nicolás' idea too, it seems to me that using parts integration after his final statement would work. $\endgroup$
    – busman
    Oct 13, 2012 at 18:40
  • $\begingroup$ 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). $\endgroup$
    – Gaspa79
    Oct 13, 2012 at 18:44
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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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  • $\begingroup$ WOW man, that's awesome!! That's waaay over my level though =P $\endgroup$
    – Gaspa79
    Oct 13, 2012 at 18:43
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    $\begingroup$ 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. $\endgroup$
    – André Nicolas
    Oct 13, 2012 at 18:52
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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 $\frac1{3x}=\tan^2u,x=\frac13\cot^2u,dx=-\frac23\cot u\csc^2udu$.

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