# Weird and difficult integral: $\sqrt{1+\frac{1}{3x}} \, dx$

My question is: How can I integrate $\sqrt{1+\frac{1}{3x}} \, dx$ ? (I don't mind about negative Xs).

I'm aware that there's a theorem which you can use that integrates the function using the points where the function can't be solved or something like that(improper integrals?) but I haven't learned anything but parts and substitution yet. I know that this integral can be solved using just those and that's what I am searching for =).

Unfortunately, this is no homework (and my calculus teacher has no idea how to solve this... I asked =( ). I just saw this integral by accident. I've been struggling with it for like a week now. I just want the answer and preferably the steps. Possible starting points:

The best approach I could find is:

$$u^2=3x \implies 2u\ du = 3\ dx$$

$$\frac{2}{3}\int \sqrt{ 1+\frac{1}{u^2} } \, \, u\ du$$

$$\frac{2}{3}\int \sqrt{ u^2+1 } \, \, du$$

Now I know that I can use sinh but I have no clue how.

That was one approach that a guy told me. If you don't like it you could start with the more traditional way:

$$u=3x$$

$$\frac{1}{3} \int\sqrt{1+\frac{1}{u}}\,du.$$ Now $$w^2=1+\dfrac{1}{u}$$ $$2w\,dw=-\frac{du}{u^2}=-(w^2-1)^2 \,du.$$ $$\frac{1}{3}\int-\frac{2w^2\,dw}{(w^2-1)^2}.$$

After that I've been told that you could integrate by parts but it's just too hard for me. I simply get nowhere. I won't type everything cause it's a waste of time. If anyone wants me to type in more work I'd be happy to if it helps in any way.

I'm in a stage where I just want to see how this super-complex (at least for me) integral is solved. I've used wolframalpha but it's nowhere near a human approach. So well... thanks a lot for any help guys! And sorry for the long post! =)!

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To call this primitive weird is uninformative and... well, weird. –  Did Oct 29 '12 at 17:52
@did well it's weird for me.... if you want I can change the title.... –  Damieh Oct 29 '12 at 18:10

For your first approach, $\int \sqrt{u^2+1} du$, you could try trig substitution, i.e. you may let $u = \tan(\theta)$, and use the identity $\tan^2(\theta)+1 = \sec^2(\theta)$, the integral after the substitution is $\int \sec^3 \theta d\theta$, to do this one you can find it on wikipedia: http://en.wikipedia.org/wiki/Integral_of_secant_cubed

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Okay after 20 minutes I realized that I am in a dead end again. After solving the integrals I end with different trigonometric functions (I used wikipedia as you said.. yay!). What should I do to replace them in form of u? Should I find the trigonometric identity between the functions I end with and the tangent? Thanks a lot! =) –  Damieh Oct 29 '12 at 19:02
I believe you find something like $\int \sec^3 xdx=\frac{1}{2}\sec x \tan x+ \frac{1}{2} \log|\sec x + \tan x|$ as in the Wikipedia article, now you know that $u = \tan \theta$, so the matter is to figure out how to represent $\sec(\theta)$ in terms of $u$, but again by the identity we have $1+u^2=1+\tan^2 (\theta) = \sec^2 \theta$ –  TTY Oct 29 '12 at 21:09
This was the way to go. By the time I read your comment I had already "solved it". I watched a khanacademy video on trigonometric substitutions and I understood how to do it. I honestly can't believe that I (well.. actually YOU) solved it. THANKS A LOT!!! –  Damieh Oct 29 '12 at 21:50

I think the following substitution is easier to work with:

$$u^2=1+\frac{1}{3x}\Longrightarrow 2u\,du=-\frac{dx}{3x^2}\Longrightarrow dx=-6u\frac{1}{9(1-u^2)^2}\,du\Longrightarrow$$

$$\Longrightarrow \int\sqrt{1+\frac{1}{3x}}dx=-\frac{2}{3}\int\frac{u^2}{(1-u^2)^2}du$$

which is already a rational integral (partial fractions and etc.).

Added $\;\;\;$ Partial fractions:

$$\frac{u^2}{(u^2-1)^2}=\frac{A}{u-1}+\frac{B}{(u-1)^2}+\frac{C}{(u+1)}+\frac{D}{(u+1)^2}\Longrightarrow$$

$$u^2=A(u-1)(u+1)^2+B(u+1)^2+C(u-1)^2(u+1)+D(u-1)^2$$

In the last polynomial identity assign values to u (recommended: $\,u=0\,,\,\pm1\,$) and compare powers of the variable (say, of $\,u^3\,$) in order to get the RHS coefficients. If I didn't make a mistake ( and I wouldn't waige on this!), one gets

$$A=B=D=\frac{1}{4}=-C$$

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Wait... is that even possible? I'm at work so I'm doing calculations on my head... Wow.. just... wow. So how can I go on from here? Should I integrate by parts? (one part $u^2$ another part $(1-u^2)^2$). Or should I try to find some substitution after cancelling the powers? Thanks a lot I'll try as soon as I go home!! =) –  Damieh Oct 29 '12 at 17:59
Read my added stuff in my answer –  DonAntonio Oct 29 '12 at 18:07
WTF? I'm guessing I'll need to learn partial fractions then. I'll read about partial fractions and watch a video or two. After that can I come back to you? =) –  Damieh Oct 29 '12 at 18:17
Ya me parecia conocido tu nick! Sos el mexicano que me ayudo cuando estaba aprendiendo limites! Jajaja muchas gracias =)!! –  Damieh Oct 29 '12 at 18:18
¿Tú sabes (o en lunfardo: ¿vos sabés?) a cuántos miles de personas he encontrado a lo largo de los años en estos sitios? Igual me da gusto el poder haberte ayudado –  DonAntonio Oct 29 '12 at 18:36