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

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?

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.

\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 $$\int\sqrt{1+\frac{1}{3x}}\ \rm{d}x=\frac{1}{2}\left[\ln(\csc\theta-\cot\theta)-\csc\theta\cot\theta\right]$$