Solve $I=\int_0^1\frac{ln(1+x)}{1+x^2}dx$
Evaluate the integral: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$
Adjust rounded rectangle radii without creating a new one in Photoshop C6 When I create new rounded rectangles, I see the radii and other information in the properties, but after I work on something else and go back to that layer, the information does not show. How do I adjust the radii of the rounded rectangles without deleting and creating a new one? zero matrix to the power of 0 Why $0^0=I$?   I'd tried prove that considering $N^0$ where N is a Nilpotent matrix and then using the Cayley -Hamilton theorem  Thanks in advance. I have a script in file bla.sh and it is executable. When I click on it, the script is executed and the window is closed. I'd like the window to stay open.  Something like command cmd /k** command in Windows.  P.S. I don't want to use pause, but I want to able to write more commands after the script was executed. Markdown in chat fails for multi-line messages macOS Catalina File Structure: Cannot remove &gt; sudo rmdir /sw rmdir: /sw: Read-only file system  &gt; mount | grep read /dev/disk1s5 on / (apfs, local, read-only, journaled)  &gt; sudo mount -o rw / &gt; mount /dev/disk1s5 on / (apfs, local, journaled) &gt; sudo rmdir /sw rmdir: /sw: Operation not permitted     root /&gt;$ mkdir a root /&gt;$ rmdir a root /&gt;$ rmdir sw &lt;b&gt;rmdir: sw: Operation not permitted&lt;/b&gt; root /&gt;$ rmdir opt &lt;b&gt;rmdir: opt: Operation not permitted&lt;/b&gt; root /&gt;$ rmdir mnt &lt;b&gt;rmdir: mnt: Operation not permitted&lt;/b&gt; root /&gt;$ rmdir home &lt;b&gt;rmdir: home: Operation not permitted&lt;/b&gt;   What exactly is preventing me from removing these directories even with SIP disabled? Can there be two distinct, continuous functions that are equal at all rationals? Akhil showed that the  is the same as the continuum, using as a step the observation that continuous functions that agree at rational points must agree everywhere, since the rationals are dense in the reals.  This isn't an obvious step, so why is it true? I have the loop tools addon enabled and when I click the circle button with these verts selected, it just disappears. Why is that?  You can see the selected options in the bottom screenshot. It is worth noting that all the verts are flat on the x plane. Locking in X or not does not change the effect.   When I use the to sphere operator it squishes the verts on Z and stretches super wide on Y, not making a cirlce.    EDIT:  here is the blend Evaluate $\lim_{n\to \infty} \sum_{r=1}^n \frac {1}{2^r}\tan \left(\frac {1}{2^r}\right)$ Evaluate $$\lim_{n\to \infty} \sum_{r=1}^n \frac {1}{2^r}\tan \left(\frac {1}{2^r}\right)$$   I tried to create a telescoping sum but I couldn't. The last step I could reach was turning the limit into $$\lim_{n\to \infty} \sum_{r=1}^n \left(\frac {1}{2^r(1-\tan (2^{-r+1})} -\frac {1}{2^r(1+\tan (2^{-r+1})}\right) $$  But couldn't proceed further. Also I thought about Riemann sums but it was a pure dead end.  Any help would be greatly appreciated Deserialize JSON object into dynamic object using Json.net Why does this steam engine fail to run? Let $a,b$ be positive integers.  When $$k = \frac{a^2 + b^2}{ab+1}$$ is an integer, it is a square.    Proof 1: (Ngô Bảo Châu): Rearrange to get $a^2-akb+b^2-k=0$, as a quadratic in $a$ this has two values: $a$ and $kb - a = (b^2-k)/a$. (The second root is determined in two different ways from the expansion $(x-r_1)(x-r_2) = x^2 - (r_1 + r_2)x + r_1 r_2$.)  Now suppose we have $a,b$ such that $k$ is an integer but not a square, by the investigation about roots we have that the second root is a nonzero integer since $k,b,a$ are integers and $k \not = b^2$, futhermore it is positive which is easily seen from its defining equation.  WLOG assume $a \ge b$ so that the second root is strictly smaller than $a$. This leads to a decent, replacing $a$ with the second root.    Proof 2 (Don Zagier): Apply reduction theory (specifically, Sätze 1 and 2 of Section 13 of my book on quadratic fields) to the quadratic form $x^2 + kxy + y^2$, which is the unique reduced quadratic form in its equivalence class.    Note that Proof 2 is pretty much the same as Proof 1 when written out in explicit detail, but I could not read Zagier's book because I cannot read German.  I would like to know more approaches to this and other alternative proofs of this result if possible! Thanks in advance.  I would also be interested in related problems (especially easier ones of a similar nature) and texts which cover the reduction theory in English. Get Record Type Developer Name without SOQL Query Custom setting Setting type not visible Multicollinearity in Decision Tree Can anybody please explain the affect of multicollinearity on Decision Tree algorithms (Classification and regression). I have done some searching but was not able to find the right answer as some say it affects it and others say it doesn't. Let $a$, $b$, and $c$ be complex numbers with $a\neq0$. Show that the solutions of $az^2+bz+c=0$ are $z_1,z_2=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, just as they are in the case when $a$, $b$, and $c$ are real numbers. Troubleshooting for biber Suppose that $(U,V)$ has the following joint pdf: $$f_{U,V}(u,v)=\exp(-\theta u-\theta^{-1}v)$$ , where $u\geq0$, $v\geq0$, $\theta&gt;0$. Define $X=UV$ and $Y=U/V$. Find the marginal pdf of $X$ and $Y$.  With transformation, I got $U=\sqrt{XY}$ and $V=\sqrt{X/Y}$ and the absolute value of Jacob is $1/2Y$. Then, the joint pdf of $(X,Y)$ is  $$f_{X,Y}(x,y)=\exp(-\theta \sqrt{XY}-\theta^{-1}\sqrt{X/Y})1/2Y, X,Y&gt;0.$$  But I am not sure how to get the marginal pdf?  Thank you! How do I format my code blocks? Free, open-source substitutes for Mendeley? Travelling as unmarried couple in Dubai