Evaluate: $\displaystyle\int\limits^{\cssId{upper-bound-mathjax}{\class{placeholder}{}}}_{\cssId{lower-bound-mathjax}{\class{placeholder}{}}} \ln\left(x+\sqrt{1+x^2}\right)\,\cssId{int-var-mathjax}{\mathrm{d}x}$   As of now, I am able to simplify it to   $\displaystyle\int\ln\left(x\right)   +  \int\ln\left(\sqrt{x^2+\dfrac{1}{x^2}}+1\right)$  Edit: The answer is  $x\ln\left(\sqrt{1+{x^2}}+x\right) - (\sqrt{1+{x^2}}) + C$
How can I integrate $$\ln \left( x+\sqrt{1+x^2} \right) $$   The problem is from IA Maron Calculus, Problem 4.3.17. The problem is under the heading 'Integration by parts'. I have tried it, but it just leads to 2 unsolvable integrals
Browse an UNC path using Windows CMD without mapping it to a network drive C:\&gt; cd \\somewhere '\\somewhere' CMD does not support UNC paths as current directories.   What I usually do to get around this is to map that directory to a network drive and then I could easily access it from the command prompt.  But is there an easier way on how to get around this? I'm using retextured snowballs and eggs for the different paintballs for two teams in a paintball map, so I can't replace the eggs with snowballs.  I have tried using /gamerule doMobSpawning false, but that still allows the chickens to spawn. I have also tried using /tp @e[type=Chicken] 0 0 0 on a repeating command block to quietly remove the chicken, but there's still a very noticeable frame where the chicken is visible. /effect @e[type=Chicken] minecraft:invisibility 10 1 false isn't fast enough, either.  I'm not using any plugins or mods.  tl;dr Without snowballs, how can I stop chickens from appearing for even a single frame. Python does not find packages installed without Anaconda What does ** (double star/asterisk) and * (star/asterisk) do for parameters? Zero visibility while editing a message in a chat room hide proof environment for restructuring Is it legal to copy Stack Overflow questions and answers? I just found http://programmingfaq.w3ec.com/faq/4761/whats-the-hi-lo-algorithm which has an exact copy of this Stack Overflow question, , with all its answers and no difference in a single character. There is no reference to Stack Overflow.  Is this legal, or at least tolerated? Is this impertinent theft of information? I just can't see any sense in copying whole pages, particularly when not referencing the source. How to Achieve this Requirement ???Can i use Jave Script , CSS and Visual force page? Imagine that you have the following list. keys = ['name', 'age', 'food'] values = ['Monty', 42, 'spam']  What is the simplest way to produce the following dictionary? a_dict = {'name': 'Monty', 'age': 42, 'food': 'spam'} Story featuring smart people sent on a space mission to nowhere The story in question featured, as the vague title suggests, a "space mission" that is really a cover for forcing a bunch of really smart people to be in close proximity to each other for long enough that they can truly evolve their collective intelligence. I recall them developing new maths and, e..g, "farseeing" and eventually figuring out that while there was no planet at the end of their journey as they were told, they were able to construct one for themselves. This is just a curiosity. I have come across multiple proofs of the fact that there are infinitely many primes, some of them were quite trivial, but some others were really, really fancy. I'll show you what proofs I have and I'd like to know more because I think it's cool to see that something can be proved in so many different ways.  Proof 1 : Euclid's. If there are finitely many primes then $p_1 p_2 ... p_n + 1$ is coprime to all of these guys. This is the basic idea in most proofs : generate a number coprime to all previous primes.  Proof 2 : Consider the sequence $a_n = 2^{2^n} + 1$. We have that  $$ 2^{2^n}-1 = (2^{2^1} - 1) \prod_{m=1}^{n-1} (2^{2^m}+1), $$  so that for $m &lt; n$, $(2^{2^m} + 1, 2^{2^n} + 1) \, | \, (2^{2^n}-1, 2^{2^n} +1) = 1$. Since we have an infinite sequence of numbers coprime in pairs, at least one prime number must divide each one of them and they are all distinct primes, thus giving an infinity of them.  Proof 3 : (Note : I particularly like this one.) Define a topology on $\mathbb Z$ in the following way : a set $\mathscr N$ of integers is said to be open if for every $n \in \mathscr N$ there is an arithmetic progression $\mathscr A$ such that $n \in \mathscr A \subseteq \mathscr N$. This can easily be proven to define a topology on $\mathbb Z$. Note that under this topology arithmetic progressions are open and closed. Supposing there are finitely many primes, notice that this means that the set $$ \mathscr U \,\,\,\, \overset{def}{=} \,\,\, \bigcup_{p} \,\, p \mathbb Z $$ should be open and closed, but by the fundamental theorem of arithmetic, its complement in $\mathbb Z$ is the set $\{ -1, 1 \}$, which is not open, thus giving a contradiction.   Proof 4 : Let $a,b$ be coprime integers and $c &gt; 0$. There exists $x$ such that $(a+bx, c) = 1$. To see this, choose $x$ such that $a+bx \not\equiv 0 \, \mathrm{mod}$ $p_i$ for all primes $p_i$ dividing $c$. If $a \equiv 0 \, \mathrm{mod}$ $p_i$, since $a$ and $b$ are coprime, $b$ has an inverse mod $p_i$, call it $\overline{b}$. Choosing $x \equiv \overline{b} \, \mathrm{mod}$ $p_i$, you are done. If $a \not\equiv 0 \, \mathrm{mod}$ $p_i$, then choosing $x \equiv 0 \, \mathrm{mod}$ $p_i$ works fine. Find $x$ using the Chinese Remainder Theorem.   Now assuming there are finitely many primes, let $c$ be the product of all of them. Our construction generates an integer coprime to $c$, giving a contradiction to the fundamental theorem of arithmetic.  Proof 5 : Dirichlet's theorem on arithmetic progressions (just so that you not bring it up as an example...)  Do you have any other nice proofs? In what cases I should skip 's in possessive case? Can "but also" be split in "not only ... but also"? I think I generally agree with the statement  When using not only . . . but also in a sentence, parallelism should be the goal. It means that the words following both parts of this correlative conjunction (i.e., not only and but also) should belong to the same parts of speech.   However, the Cambridge dictionary has an example that is not 100% parallel in my eyes, as &quot;but also&quot; is split:  Not only did she forget my birthday, but she also didn’t even apologise for forgetting it.   What I would expect, with maximum parallelism, is  Not only did she forget my birthday, but also she didn’t even apologise for forgetting it.  Is my expectation wrong? Is my formulation acceptable? Why Integer.getInteger does not work? Huge paged pool probably caused by fltMgr.sys , what to do? How to deal with cross-thread access exceptions? Xdotool using "DISPLAY=:0" not works in Crontab Why can't I launch eclipse from a terminal ? change $PATH environment variable In Ubuntu 18.04.1, I would like to use my left Super-key only to simply open the Show Applications menu (aka Dash) like a left click on the dock-icon (with the 9 dots) does and preferably close it again without showing me any open programs.   I can open it with Super + A and even close it again with Super R, but Super L appears to do nothing at all.   In Xfce, something similar was possible by adding a custom keyboard shortcut for Super L and setting the command to xfce4-popup-whiskermenu, which would open and close a menu with favourite applications. Unfortunately, I don't know the command that opens and closes the Show Applications menu.   Is there any way to add such a shortcut in the new Gnome Shell? \documentclass{article} \usepackage{algorithm}  \begin{document} \setcounter{section}{54} \section{A section with no label} \begin{algorithm} \label{myalg} \caption{An algorithm with the label `myalg'.} \end{algorithm} Reference to myalg: \ref{myalg}. \end{document}   With this code, the reference givens 55 (the section number); if I move the label after the caption command, I get the correct reference (1).  Why should this be the case? I find it confusing: when thinking about an algorithm as an environment, you expect to be able to label it right after you began it. At the very least I would expect some kind of error or "??" to appear, rather than the previous environment or reference-able entity being used.