Sometimes I see that, during a $u$-substitution in an integral (which is something I hardly understand), once $u$ is chosen it is differentiated to obtain $du/dx$. I understand this. After, however, people solve for $dx$ as though it is a regular variable, and treat $du/dx$ as though it is a regular fraction. My googling has told me that $du/dx$ isn't a fraction and that $dx$ or $du$ aren't really regular variables, as one is dependent on the other. So why can they behave like what they are not?  For example, if $u$ is $x^2$, then $du/dx$ ends up being $2x$, and so $du = 2x dx$ and $dx = du/2x$. What my question is, why is this allowed?  (Also, sorry that I haven't properly formatted the previous equations using latex or whatever it should be, as i don't know about any of that right now.)
7365
In the book Thomas's Calculus (11th edition) it is mentioned (Section 3.8 pg 225) that the derivative $dy/dx$ is not a ratio. Couldn't it be interpreted as a ratio, because according to the formula $dy = f'(x)dx$ we are able to plug in values for $dx$ and calculate a $dy$ (differential). Then if we rearrange we get $dy/dx$ which could be seen as a ratio.   I wonder if the author says this because $dx$ is an independent variable, and $dy$ is a dependent variable, for $dy/dx$ to be a ratio both variables need to be independent.. maybe?
82555
What is $\int{x^{x^{x^{x^{\dots }}}}dx}$? I managed to get it down to a complicated series after using $u$-substitution, but I want to see other methods of approach, or if my answer was right.   First I let $$u=x^{x^{x^{x^{\dots }}}}$$, and $du$ would be, in terms of $u$, $$du=  \frac{u}{\frac{1-ln(u)}{u}e^{ln(u)/u}}$$ This makes the integral of $$\int{x^{x^{x^{x^{\dots }}}}dx}=\int \frac{1 - ln(u)}{u}{e^{ln(u)/u} }du$$ Now making $w = ln(u), dw$ becomes $1/u$ and the integral becomes $$\int e^{w/e^{w}}(1-w)dw$$ From there, I distributed and made $2$ integrals of series, using the Taylor polynomial for $e^x$. From there, I integrated and back substituted. I was only trying to find the indefinite integral.   I'm still a junior in high school, so let me know if the answer is beyond my scope of understanding.
8817
$dx$ appears in differential equations, such us derivatives and integrals.  For example, a function $f(x)$ its first derivative is $\dfrac{d}{dx}f(x)$ and its integral $\displaystyle\int f(x)dx$. But I don't really understand what $dx$ is.
89637
I know it is defined as $ dP \wedge dx  $ or explicitly,  $$ \frac{\partial P }{\partial y } dy \wedge dx   . $$  The question is, could it be $dx \wedge d P $?  Or  $$  \frac{\partial P }{\partial y }  dx \wedge dy ?    $$  I know the question must be very naive or even stupid. But I am indeed confused.
84672
Letting $x=\tan(u)$, $$\int\frac{1}{1+x^2} \, dx=\int\frac{1}{1+\tan(u)^2}\sec^2(u) \, du=u+k=\arctan(x)+k$$ Also,  $$\int\frac{1}{1+x^2} \, dx=\int\frac{1}{(x-i)(x+i)} \, dx=\frac{1}{2i}\int\frac{1}{(x-i)}-\frac{1}{(x+i)} \, dx$$ $$=\frac{1}{2i}\left(\ln(x-i)-\ln(x+i)\right)+c$$ $$=\frac{1}{2i}\ln \left(\frac{x-i}{x+i} \right)+c$$ Giving $$\arctan(x)=\frac{1}{2i}\ln \left(\frac{x-i}{x+i} \right)+q$$  Why is this correct? What is the nature of $q$ (is it 'flexible' so the equality doesn't really mean much)?  I think it probably has something to do with the relationship between $\log(z)$ and $\arg(z)$, but $\arg(z\pm i)$ is hard to calculate neatly.
55214
The Problem is to integrate the following double integral $\displaystyle\iint_\mathbf{D}\exp\left(\frac{x-y}{x+y}\right) dx\, dy$ using the technique of transformation of variables ($u$ and $v$).  There is a given $D$ with $D:= \{(x,y):\ x\geq 0,\ y \geq 0,\ x+y \leq 1\}$  My Approach: wasn't quite successful at all. I am quite familar with calculus, but i am stuck at multiple integrals. I thought about the following transformation:  $u = x-y$ and $v = x+y$  But i don't know the next step.  Another suggestion: i thought about the integration of the e-Function. I have in mind that $\int e^x dx = e^x$. So maybe: $$\iint_\mathbf{D} \exp\left(\frac{x-y}{x+y}\right) dx\ dy = \exp\left(\frac{x-y}{x+y}\right) ?$$
100750
Why is it when evaluating the definite integral we commonly opt to omit the constant $c$ $$\int_1^2x^2 \, dx= \left.\frac{x^3}{3} \right|_1^2 =\frac{2^3}{3}-\frac{1^3}{3}=\frac{7}{3}$$ But when evaluating the indefinite integral we include it? $$\int x^2dx = \frac{x^3}{3} + c$$ Would it not be more accurate to instead do the following? $$\int_1^2 x^2\,dx=\left.\frac{x^3}{3}+c\right|_1^2 = \left(\frac{2^3}{3}+c\right) - \left(\frac{1^3}{3}+c\right) = \left(\frac{7}{3}+c\right)$$ Is there a historical reason to all of this?
6310
I've always wondered why the numerator is $d^n$ while the denominator is $dx^n$ instead of $d^nx$ like the numerator.  I must be missing something very obvious or fundamental. Is this notation derived from the chain rule in some way?
21347
There is supposed to be a clean solution to the integral below, maybe involving some symmetry $$ \int^1_0 \frac{\ln(1+x)}{x}dx$$ I have tried integration by parts as followed: $\ln(x+1)=u$ ,$\frac{1}{x+1} dx = du$ and also $\frac{1}{x}dx=dv$, $\ln(x)=v$. Then, the integral becomes $$\int^1_0 \frac{\ln(1+x)}{x}dx=- \int^1_0 \frac{\ln x}{1+x}dx$$ which does not make this easier. I have also tried using the identity $\int_a^bf(x)~dx=\int_a^bf(a+b-x)~dx$. So let $I = \int^1_0 \frac{\ln(1+x)}{x}dx$ and also $I = \int^1_0 \frac{\ln(2-x)}{1-x}  $. Then $$2I= \int^1_0 \ln(1+x)+\ln(2-x)$$ which is not any easier, either. Any ideas? :)
19965
Evaluate the PDF to find K  $1 = \int_0^{\infty} \frac{k}{(1+x)^4}$  using u substitution I got:  $u=1+x$ and $du= dx$  $\int_0^{\infty} \frac{k}{u^4} \to \frac{k}{\frac{-1}{3} u^3}$   Since $u^4 du = \frac{u^{-4+1}}{-4+1}=$ $\frac{u^{-3}}{-3}$  which evaluating gives:  $1=\frac{k}{\frac{-1}{3(1+x)^3}} \vert_0^{\infty} \to \frac{-3k}{(1+x)^3} \vert_0^{\infty} = 0 -3k$   and $k= \frac{-1}{3}$  somehow though they came up with $\frac{-k}{3} \frac{1}{(1+x)^3}$ after integrating $\int_0^{\infty} \frac{k}{(1+x)^3}$   Was my approach correct to use substitution? Where does my work go wrong?
117830
How to integrate $\frac{1}{x\log x}$?  Could you give me some ideas on how to integrate this? thanks.  i've tried setting $u=(\log x)^{-1}$.   $\dfrac{\mathrm du}{\mathrm dx} = x^{-1}$  But it didnt work...
105361
This is about that old chestnut, $\newcommand{\d}{\mathrm{d}} \int \sin^2 x\,\d x$.   OK, I know that ordinarily you're supposed to use the identity $\sin^2 x = (1 - \cos 2x)/2$ and integrating that is easy. But just for the heck of it, I tried using the $u$-$v$ substitution method (otherwise known as integration by parts).  $$ \int \sin^2 x\,\d x = \int \sin x \sin x\,\d x $$  We can say $u=\sin x$ and $\d u=\cos x\,\d x$ while $\d v = \sin x\,\d x$ and $v = -\cos x$. When we put it all together:   $$\int \sin^2 x\,\d x = u v - \int v\,\d u = -\sin x \cos x - \int -\cos^2 x\,\d x$$  and doing the same routine with $\cos^2 x$ we get $u = \cos x$, $\d u = -\sin x\,\d x$, $\d v = \cos x\,\d x $ and $v = -\sin x$, leading to:  $$\begin{align}   \int \sin^2 x\,\d x = uv - \int v\,\d u     &amp;= -\sin x \cos x - (-\cos x \sin x - \int -\sin^2 x\,\d x) \\     &amp;= -\sin x \cos x + \sin x \cos x - \int \sin^2 x\,\d x \end{align}$$  which eventually works out to  $$2\int \sin^2 x\,\d x = 0$$  So I wanted to get an idea why this didn't work. Maybe it's higher math and the why will be beyond me (I would think that might be the case), or maybe it's one of those proofs that looks absurdly simple when shown that I am just unaware of.
24011
We are asked to prove the following integral;  $\int_0^x(\int_0^tf(u) du) dt=\int_0^xf(u)(x-u)du$   I know that we have to use integration by parts which when we differentiate a function will help us use the fundamental theorem of calculus. I just don't know when and were to apply this.  Thanks in advance!
34161
The question I am working on: Evaluate   $$\frac{1}{2} \int^1_0{x^4 (1-x)^4 } dx \le \int^1_0{\frac{x^4 (1-x)^4}{1+x^2}} dx \le \int^1_0{x^4 (1-x)^4 } dx$$  So using integration by parts to solve:   (letting $u=(1-x)^4$ and $dv=x^4$)  $$\int{x^4 (1-x)^ 4} dx = \frac{4}{5}x^5(x-1) - \frac{4}{5}\left(\frac{x^7}{7} - \frac{x^6}{6}\right)+c$$  Is it correct so far? If so ...   $$\int^1_0{ x^4 (1-x)^4 } dx = \frac{4}{5}\left(\frac{1}{6}-\frac{1}{7}\right)=\frac2{105}$$  $$\frac{1}{2} \int^1_0{ x^4 (1-x)^4 } dx = \frac{2}{5}\left(\frac{1}{6}-\frac{1}{7}\right)=\frac1{105}$$  But   $$\int\frac{{x^4(1-x)^4}}{1+x^2} dx = ??$$    Since I found $\int{{x^4(1-x)^4}} dx$. I thought of integration by parts, letting $u=\frac{1}{1+x^2}$, $dv = x^4(1-x)^4$. But I will get a very complicated $v$ to integrate later? Same if I did it the other way around?
42702
Of course, it is easy to see, that the integral (or the antiderivative) of $f(x) = 1/x$ is $\log(|x|)$ and of course for $\alpha\neq - 1$ the antiderivative of $f(x) = x^\alpha$ is $x^{\alpha+1}/(\alpha+1)$.  I was wondering if there is an intuitive (probably geometric) explanation why the case $\alpha=-1$ is so different and why the logarithm appears?  Some answers which I thought of but which are not convincing:   Taking the limit $\alpha=-1$ either from above or below lead to diverging functions. Some speciality of the case $\alpha=-1$ are that both asymptotes are non-integrable. However, the antidrivative is a local thing, and hence, shouldn't care about the behavior at infinity.
96462
I think I'm having a bad day. I was just trying to use integration by parts on the following integral:  $$\int \frac{dx}{x\log(x)}$$ Which yields $$\int \frac{dx}{x\log(x)} = 1 + \int \frac{dx}{x\log(x)}$$ Now, if I were to subtract  $$\int \frac{dx}{x\log(x)}$$ from both sides, it would seem that $0 = 1$. What is going on here?  Aside: I do know the integral evaluates to  $$\log(\log(x))$$ plus a constant.
22772
Given $f = f(u(t), t)$, is it possible to find the following derivative? ($f$,$u \in \mathbb{R}^n$, $t \in \mathbb{R}$)  $$ \frac{\partial f}{\partial||u||^2_2} $$  I am aware of the following derivatives:  $$ \frac{\partial||x||^2_2}{\partial x} = \frac{\partial ||x^Tx||_2}{\partial x} = 2x $$  So I can invert it to give me (elementwise division in the RHS)  $$ \frac{\partial x}{\partial||x||^2_2} = \frac{1}{2x} $$  This step does feel like abuse of notation I find it hard to convince myself of its validity.  If it's right though, I could just apply the chain rule to find  $$ \frac{\partial f}{\partial u} \frac{\partial u}{\partial||u||^2_2} $$  However, knowing that both $f$ and $u$ are a function of $t$, I am somehow doubting my own reasoning. Or does that just mean that the derivative will still be a function of $t$ and I have nothing to worry about? Can anyone shed some light?  UPDATE: I'm now doubting the existence of  $$ \frac{\partial u}{\partial ||u||^2_2} $$  because $u = u(||u||_2^2)$ is not injective. Is the inversion step I did above legitimate? Is there another possible approach?