Evaluate $\int_0^1 \frac{(x^2-1)}{\log x} dx$
What is $\int_0^1\frac{x^7-1}{\log(x)}\mathrm dx$?
I'm try to change to my Desktop directory from the terminal.  Running  cd /home/desktop   gives an error in the terminal saying no directory found.  What am I doing wrong? How can I count the number of records that have a unique value in a particular field in ROR? Does Find My Mac stop working after the system has been wiped? Can I tap a creature with summoning sickness to pay "tap an untapped creature you control" costs? I tried to install Ubuntu 12.10 with the Wubi installer while running Windows, because my PC does not accept my BootCD as a bootable medium.  But Wubi showed me this error:  Ubuntu-Installationsprogramm (X) Ein Fehler ist aufgetreten: Could not retrieve the required disk image files Weiter Informationen finden Sie in der Protokolldatei: c:\users\username\appdata\local\temp\wubi-12.10-rev273.log   What can I do, because I want to get rid of my Windows 7 and use the way better Ubuntu 12.10?  Thanks for your huge answer, now my next problem after trying to install wubi with the .iso is, that the same error shows up again and i am not shure which part of the errorreport you need.  Can you tell me? Because i'm not that good at linux. I am looking for a word that would describe being obsessed with the details of a larger entity such that the "looker" neglects to see the whole or (perhaps more importantly) the purpose of the whole.  Basically, "not seeing the big picture".  What can [more] succinctly describe this?  Edit: I just found this post: . I think I want the exact opposite. Let $q_1,q_2,q_3,q_4$ be orthonormal vectors in $\mathbb{R}^4$,$z_1,z_2,...,z_6$ orthonormal vectors in $\mathbb{R}^6$ and $A=z_1q_1^T + z_2q_2^T$. Let $q_1,q_2,q_3,q_4$ be orthonormal vectors in $\mathbb{R}^4$,$z_1,z_2,...,z_6$ orthonormal vectors in $\mathbb{R}^6$ and $A=z_1q_1^T + z_2q_2^T$.  a) Find the base and dimension of fundamental subspaces of matrix A  b) Using the results given by a) find the general solution for $Ax - z_1$.  Any hint would be appreciated. I’m reading Shakespeare’s Henry IV (Part 1) right now and I’m noticing that Sir John Falstaff has a propensity of saying “If this, then I’m a this” sort of statements. A few examples to clarify:     … An I   have not ballads made on you all and sung to filthy   tunes, let a cup of sack be my poison      An I have not   forgotten what the inside of a church is made of, I   am a peppercorn, a brewer’s horse      if manhood, good manhood, be   not forgot upon the face of the earth, then am I a shotten herring      If I do not beat thee out of thy   kingdom with a dagger of lath, and drive all thy   subjects afore thee like a flock of wild-geese,   I’ll never wear hair on my face more   Is there a particular literary term for these sort of sentences? Let $A$ be a UFD. Assume that $a,b \in A$ are relatively prime, $c \in A$ and $a | bc$. To prove that $a|c$, is the following approach correct (or do you have to use some type of prime factorization argument)?  By the relatively prime assumption, $\gcd(a,b)$ is a unit. Call that unit $u \in A^{\times}$. Then we can write $u = ax + by$ for some $x,y \in A$. But then, $c = c \cdot u = c(ax+by)=cax+cby$. However, it is clear that $a|cax$ and the fact that $a|bc$ implies $a|cby$. Thus, $a|cax+cby=c$. I'm trying to find a copy of 6.20 to run a diff on a client's Drupal install. I believe there's been changes to core.   How can I get version 6.20? Additionally, how can I download any older Drupal versions? How can I create a new extensible symbol? Given an arbitrary thin vertical symbol, can I create a new math delimiter from it (and if so, how)?   Edit: Ideally, I would like to be able to create an extensible symbol, either built out an existing normal symbol, or even made from scratch.  Edit 2: I am pretty sure that what I am after is an explanation of how to create a .tfm file for an extensible symbol (which Werner's answer describes the general structure of, but not how to make one).    For example, let's say I wanted to use the \dagger () or \wr () symbol as a delimiter, and write expressions like    \[\left\ldagger \sum_{n=1}^\infty a_n \right\rdagger\]   where the symbol scaled appropriately. Would there be any distinction between declaring a symmetric delimiter (like \vert), as opposed to one that has inherently different "left" and "right" versions like [ and ]?  Now, even being unaware of the inner workings of TeX, I imagine that there has to be lots of information in the fonts for how a delimiter is to be scaled in different situations, and if I try to make my own delimiter, this information won't be present. Would I need to get deep into FontForge or equivalent to achieve my custom delimiters, or is there an easier way? SCJP6 regex issue Do mankind and manmade activities/constructions have any effect on the rotation of the Earth? We walk or ride on our vehicles to our destinations daily. Does our movement have any effect on the rotation of the earth according to Newton's law? What will be the effect if we move all the peoples along with their vehicles at their maximum velocity in one line in one direction along the equator? How much effect will it make? Let $f:\mathbb{R}_+ \to \mathbb{R}_+$ be a monotone decreasing function defined on the positive real numbers with $$\int_0^\infty f(x)dx &lt;\infty.$$ Show that $$\lim_{x\to\infty} xf(x)=0.$$   This is my proof: Suppose not. Then there is $\varepsilon$  such that for any $M&gt;0$ there exists $x\geq M$ such that $xf(x)\geq \varepsilon$. So we can construct a sequence $(x_n)$ such that $x_n \to \infty $ and $x_n f(x_n ) \geq \varepsilon$. So $$\frac{\varepsilon}{x_n}\leq f(x_n) \implies \sum_{n\in\mathbb{N}}\frac{\varepsilon}{x_n} \leq  \sum_{n\in\mathbb{N}} f(x_n) \leq \int_0^1 f(x)dx.$$ So we get a contradiction.  I feal like I have the correct idea but some details are wrong.  Any help would be appreciated. I killed my first wife once I found out there is a mercenary you can marry who gives you up to 300 gold a day (in Whiterun). The only requirement, of course, being that I had to hire her first. So I hired her, and then I popped the question, to which she replied the usual: tell the priest to arrange the wedding.   When we reached Riften I told her to wait, just outside the bee and barb, and arranged the wedding. I went through the process, and the second it finished she left, I followed but as I left she was nowhere to be seen. I have tried going back to my house, the pub where I hired her and that same spot in Riften about 6 times each now, some during the day, some during the night and cannot find her at all.   Now this isn't to much of a problem because I can re-marry...oh wait, to remarry, my spouse has to die, so I can't claim gold off her and can't remarry, and the autosaves have been used up fast travelling to the places I thought she might've been. My last save is something like 6 levels ago, and I don't want to have to load it.   What should I do about this situation? Why are my photos not crisp? I use a Canon EOS Rebel T3i (600D) with the kit lens. I have experimented a lot with different settings and modes, but I feel like I have wasted many good shots that could have looked amazing, if only they were a little bit sharper.  I use auto-focus on most of my pictures and make sure to choose the right focus point, but all of my photos are still a little bit out of focus. When I zoom in, I can see that the pixels are not very crisp or well-aligned.  I am not sure if this issue is caused by my equipment or skills. Underline my name in the bibliography For several documents I'm preparing, I need to highlight (or underline) occurences of my name within the bibliography.   Is there an easy way of doing this? Non-hidden directory doesn't show in Explorer, but the program that made it can access it How to find $\sum_{r\ge 0} \binom{n}{r}\binom{n-r}{r} 2^{n-2r}$? Problem was to find $$\sum_{r\ge 0} \binom{n}{r}\binom{n-r}{r} 2^{n-2r}.$$ My partial progress i tried to motivate such that upper term in binomial terms gets constant rather than variable , so i thought of changing it to form $\binom{n}{n-r}\binom{n-r}{n-2r}$. Now it's of the form $\binom{n}{m}\binom{m}{k}$. We know it is equivalent to $\binom{n}{k}\binom{n-k}{m-k}$. Doing this we get $\binom{n}{2r}\binom{2r}{r}2^{n-2r}$. Now what I need to do as next step as such still nothing the upper or lower terms r any constant values(sum to ) , all are variables still. Now by one answer by Robpratt i got by snake oil , is there any other way to solve this problem ? Why does my grep expression need to use $'string' to match tab characters?