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! =)!
 A: \begin{equation}
\int\sqrt{1+\frac{1}{3x}}\ dx
\end{equation}
Let
\begin{align*}
x&=\frac{\cot^2\theta}{3},\\
\frac{\rm{d}x}{\rm{d}\theta}&=\frac{2}{3}(\cot\theta)(-\csc^2\theta)
\end{align*}
Use the above substitution in (1)
\begin{equation}\begin{split}
 -\frac{2}{3}\int\sqrt{1+\tan^2\theta}\cot\theta\csc^2\theta \ d\theta &=-\frac{2}{3}\int\sec\theta\cot\theta\csc^2\theta \ d\theta\\
 &=-\frac{2}{3}\int\frac{\cot\theta}{\cos\theta\sin^2\theta}\ \rm{d}\theta\\
 &=-\frac{2}{3}\int\csc^3\theta \ d\theta
\end{split}\end{equation}
From here, we need integration by parts
\begin{align*}
u&=\csc\theta,\quad &dv=\csc^2\theta \ d\theta,\\
\rm{d}u&=-\cot\theta\csc\theta \quad &v=-\cot\theta
\end{align*}
We now have
\begin{equation*}
\begin{split}
-\frac{2}{3}\int\csc^3\theta \ \rm{d}\theta&=-\frac{2}{3}\left[-\csc\theta\cot\theta-\int\cot^2\theta\csc\theta \ d\theta\right]\\
&=-\frac{2}{3}\left[-\csc\theta\cot\theta-\int(\csc^2\theta-1)\csc\theta \ d\theta\right]\\
&=-\frac{2}{3}\left[-\csc\theta\cot\theta-\int\csc^3\theta\ d\theta +\int\csc\theta)\ \rm{d}\theta\right]\\
&=-\frac{2}{3}\left[-\csc\theta\cot\theta-\int\csc^3\theta\ d\theta +\int\csc\theta\left(\frac{\csc\theta-\cot\theta}{\csc\theta-\cot\theta}\right)\ d\theta\right]\\
&=-\frac{2}{3}\left[-\csc\theta\cot\theta-\int\csc^3\theta\ d\theta +\int\frac{\csc^2\theta-\csc\theta\cot\theta}{\csc\theta-\cot\theta}\ d\theta\right]
\end{split}
\end{equation*}
Now, in the rightmost integral we make the substitution $$u=\csc\theta-\cot\theta, \quad\rm{d}u=(\csc^2\theta-\csc\theta\cot\theta)\ d\theta$$
and we end up with
\begin{equation*}
-\frac{2}{3}\int\csc^3\theta\ d\theta=\frac{2}{3}\csc\theta\cot\theta+\frac{2}{3}\int\csc^3\theta\ \rm{d}\theta-\frac{2}{3}\ln(\csc\theta-\cot\theta)
\end{equation*}
Finally collecting similar terms we get
\begin{equation*}
\begin{split}
-\frac{4}{3}\int\csc^3\theta\ d\theta&=\frac{2}{3}\csc\theta\cot\theta-\frac{2}{3}\ln(\csc\theta-\cot\theta),\\
\int\csc^3\theta\ d\theta&=-\frac{1}{2}\csc\theta\cot\theta+\frac{1}{2}\ln(\csc\theta-\cot\theta),
\end{split}
\end{equation*}
So the final result is
$$\int\sqrt{1+\frac{1}{3x}}\ \rm{d}x=\frac{1}{2}\left[\ln(\csc\theta-\cot\theta)-\csc\theta\cot\theta\right]$$
A: 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
A: 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
$$A=B=D=\frac{1}{4}=-C$$
A: Another strategy could be
$$\int\sqrt{1+\frac{1}{3x}}dx=\int\sqrt{\frac{1}{3x}\left(1+3x\right)}dx=\frac{2}{\sqrt3}\int\frac{1}{2\sqrt{x}}\left(\sqrt{1+3x}\right)dx$$
Now let $\sqrt{x}=y$, hence $\frac{1}{2\sqrt{x}}dx=dy$
$$\frac{2}{\sqrt3}\int\frac{1}{2\sqrt{x}}\left(\sqrt{1+3x}\right)dx=\frac{2}{\sqrt3}\int\sqrt{1+3y^2}dy$$
Now let $\sqrt{3}y=\sinh z$, hence $dy=\frac{\cosh z}{\sqrt3}dz$
$$\frac{2}{\sqrt3}\int\sqrt{1+3y^2}dy=\frac{2}{3}\int(\cosh z)^2dz=\frac{1}{6}\int(e^z-e^{-z})^2dz=\frac{1}{12}e^{2z}-\frac{1}{12}e^{-2z}+\frac{1}{3}z+c$$
Finally we have
$$\int\sqrt{1+\frac{1}{3x}}dx=\frac{1}{12}e^{2arcsinh(\sqrt{3x})}-\frac{1}{12}e^{-2arcsinh(\sqrt{3x})}+\frac{1}{3}arcsinh(\sqrt{3x})+c$$
A: This is a nice candidate for the “integral of the inverse” technique.
The idea is to use the integration by parts formula in a weird way:
$$\int {y dx} = xy - \int{x dy}$$
Let $ y= \sqrt{1+ \frac{1}{3x} } $.
So $x=\frac{1}{3} \frac{1}{y^2-1} $
Plugging this into the integration by parts formula:
$$ \int{\sqrt{1+\frac{1}{3x}}dx} = x\sqrt{1+\frac{1}{3x}} - \frac{1}{3}\int {\frac{dy}{y^2-1}} $$
which is much easier to integrate.  Afterwards, substitute $ y= \sqrt{1+ \frac{1}{3x} } $ for any remaining y’s.
A: Method 1: Let's try integration by parts
\begin{eqnarray*}
\int\sqrt{u^2+1}\,du&=&\int1\cdot\sqrt{u^2+1}\,du=u\sqrt{u^2+1}-\int u\frac{2u}{2\sqrt{u^2+1}}\,du=\\
&=&u\sqrt{u^2+1}-\int \frac{u^2+1-1}{\sqrt{u^2+1}}\,du=
u\sqrt{u^2+1}-\int\left(\sqrt{u^2+1}-\frac{1}{\sqrt{u^2+1}}\right)\,du=\\
&=&u\sqrt{u^2+1}-\underbrace{\int\sqrt{u^2+1}\,du}_{\text{same}}+\int\frac{1}{\sqrt{u^2+1}}\,du.
\end{eqnarray*}
The unknown integral can be solved now as
$$
2\int\sqrt{u^2+1}\,du=u\sqrt{u^2+1}+\int\frac{1}{\sqrt{u^2+1}}\,du
$$
which gives you the answer if you know how to calculate the last integral (quite standard). If you don't then you can try
Method 2: The standard substitution for "square root of the square plus constant" 
$$
t=u+\sqrt{u^2+1}
$$ 
that gives after some algebra 
$$
u=\frac{t^2-1}{2t}=\frac{t}{2}-\frac{1}{2t},\quad du=\frac{t^2+1}{2t^2}\,dt,\quad  \sqrt{u^2+1}=t-u=\frac{t^2+1}{2t}
$$
$$
\int\sqrt{u^2+1}\,du=\int\frac{(t^2+1)^2}{4t^3}\,dt=\int\frac{t^4+2t^2+1}{4t^3}\,dt=\frac{1}{4}\int\left(t+\frac{2}{t}+\frac{1}{t^3}\right)\,dt
$$
which I believe you can manage yourself.
P.S. When doing the final substitution back to $u$ it is beneficial to note that
$$
\frac{1}{t}=\frac{1}{\sqrt{u^2+1}+u}=\frac{1}{\sqrt{u^2+1}+u}\cdot\frac{\sqrt{u^2+1}-u}{\sqrt{u^2+1}-u}=\frac{\sqrt{u^2+1}-u}{u^2+1-u^2}=\sqrt{u^2+1}-u.
$$
P.P.S. Can you now calculate the last integral in the method 1 by the method 2?
A: You can indeed integrate by parts.
\begin{align*}
& \text{Let } \: t =\sqrt{1+\frac{1}{3x}}. \text{ Then, } x = \frac{1}{3} \frac{1}{(t^2-1)}\text{ so } dx = \frac{2t}{3(t^2-1)^2}\: dt \\
\\
& \text{So substituting to our integral we get that, }\\
\\
&\int_{}^{}\sqrt{1+\frac{1}{3x}}dx = \int_{}^{} t\frac{2t}{3(t^2-1)}dt = \frac{1}{3} \int_{}^{}\frac{2t^2}{(t^2-1)^2}dt= \frac{1}{3}\int_{}^{}(\frac{1}{t^2-1})'t\:dt=\\
&\frac{1}{3}(\frac{t}{t^2-1} - \int_{}^{} \frac{1}{t^2-1}dt)=\frac{1}{3}(\frac{t}{t^2-1} - \frac{1}{2}\int \frac{1}{t-1}- \frac{1}{t+1}\:dt)=\\
&\frac{1}{3}[\frac{t}{t^2-1} - \frac{1}{2}(\ln \frac{t-1}{t+1}) ] + c = \frac{1}{3}(\frac{t}{t^2-1}) - \frac{1}{6}\ln (\frac{t-1}{t+1}) + c\\
\\
&\text{Finally, by substituting x back again (and symplifying a bit) we get, }\\ \\
&x\sqrt{1+\frac{1}{3x}} + \frac{1}{6}\ln{[x(\sqrt{1+\frac{1}{3x}}+1)^2]} + c
\end{align*}
