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9

hint: $\dfrac{f(x^2-4)}{x-2} = (x+2)\cdot \dfrac{f(x^2-4)}{x^2-4}$, and then use the result above.


6

Differentiating both the expression $$\frac{1}{4} \text{arcsec } \frac{x^2}{2} + C$$ derived in the question and the expression $$-\frac{1}{4} \arctan \frac{2}{\sqrt{x^4 - 4}} + C$$ given by WolframAlpha yields the original integrand, so both are correct. We can use a little easy trigonometric to see how this can be: At least when $u \geq a > 0$, the ...


5

HINT: Set $e^x=u, x=\ln u\implies dx=\dfrac{du}u$ $$\int\dfrac{dx}{e^x(e^x+1)}=\int\dfrac{du}{u^2(u+1)}$$ Now use Partial fraction decomposition, $$\dfrac1{u^2(u+1)}=\dfrac Au+\dfrac B{u^2}+\dfrac C{u+1}$$


5

Integrating $f(z)=\dfrac{e^{iaz^2}}{z+b}$ along $[0,R]\cup Re^{i[0,\pi/2]}\cup i[R,0]$ gives \begin{align} \int^\infty_0\frac{\sin(ax^2)}{x+b}\ {\rm d}x &=\int^\infty_0\frac{b\cos(ay^2)-y\sin(ay^2)}{y^2+b^2}\ {\rm d}y \end{align} To compute the first integral, we consider the function $\displaystyle I(a)=\int^\infty_0\frac{e^{iay^2}}{y^2+b^2}\ {\rm d}y$ ...


3

Ok, i will give it a shot: Writing $\int_{0}^{\infty}e^{-t(x+2)}=\frac{1}{x+2}$ and using $\Im(e^{ix})=\sin(x)$ we may reformulate the problem as follows: $$ I=\Im\left[\int_0^{\infty}dte^{-2 t}\underbrace{\int_0^{\infty}dxe^{i\pi x^2-tx}}_{J(t)}\right] $$ the inner intgral $J(t)$ is quite straightforward (and also well known because it is just the laplace ...


3

Write $$\dfrac1{e^x(e^x+1)}=\dfrac{(e^x+1)-e^x}{e^x(e^x+1)}=e^{-x}-\dfrac1{e^x+1}$$ and use How do I solve $\displaystyle\int \frac{\mathrm{d}x}{e^x + 1} $?


3

And yet, one more approach is $$\begin{align} \int \frac{1}{e^{2x}+e^x}\,dx&=\int \frac{e^{-2x}}{1+e^{-x}}\,dx\\\\ &=\int \left(1-\frac{1}{1+e^{-x}}\right)e^{-x}\,dx\\\\ &=-e^{-x}+\int \frac{1}{1+e^{-x}}\,d\left(e^{-x}\right)\\\\ &=\log(1+e^{-x})-e^{-x}+C \end{align}$$


3

W.l.o.g. assume $a_n=1$. Then by Vieta's formulas we have $-a_{n-1}=x_1+x_2+\dotsc+x_n, a_{n-2}=x_1x_2+x_1x_3+\dotsc+x_{n-1}x_n, \dotsc, (-1)^na_0=x_1x_2\dotsc x_n$ These are called the elementary symmetric polynomials. Now, you might observe that $$x_1+x_2+\dotsc+x_n=-a_{n-1}$$ $$x_1^2+x_2^2+\dotsc+x_n^2=a_{n-1}^2-2a_{n-2}$$ Furthermore, one might obtain ...


3

Let $\sigma_i(x_1,\ldots,x_n)$ be the $i$th order basic symmetric polynomial of $x_1,\ldots,x_n$. Since: $$ a_{n-i}=\sum_{1\leq k_1<\ldots<k_i\leq n} \prod_{j=1}^i -x_{k_j}=:\sigma_i(-x_1,\ldots,-x_n) $$ And by the fundamental theorem of the symmetric polyinomials, there is a polynomial such that: $$ \sum_{i=1}^n ...


3

The idea is to factor out the dominating term: The limit as $x\to+\infty$: If $x>0$, $$ \frac{\sqrt{x^2-1}}{2x+1}=\frac{x\sqrt{1-1/x^2}}{x(2+1/x)}=\frac{\sqrt{1-1/x^2}}{2+1/x}\to\frac{\sqrt{1}}{2}=\frac{1}{2}, $$ where the limit is taken as $x\to+\infty$. If you want to study the limit as $x\to-\infty$, you should be a bit more careful, and factor out ...


3

If $a\in\mathbb{N}$, factor $1+t^a$ then exploit: $$ \int \frac{\log(1-\alpha t)}{1+t}\,dt = \log(1-\alpha t)\log\left(\frac{\alpha+\alpha t}{1+\alpha}\right)+\text{Li}_2\left(\frac{1-\alpha t}{1+\alpha}\right). $$ The last line can be easily checked through differentiation.


3

$\bullet $ Integration is Reverse process of Differentiation. So we know that $\displaystyle \frac{d}{dx}(\tan x+\mathcal{C}) = \sec^2 x\;,$ Now Integrate both side w r to $x$ So $$\displaystyle\int \frac{d}{dx}(\tan x+\mathcal{C})dx = \int \sec^2 x dx$$ So $$\displaystyle \tan x+\mathcal{C}=\int \sec^2 xdx = \int\frac{1}{\cos^2 x}dx$$ So $$\displaystyle ...


2

One way to do this is by the product rule for a scalar valued and vector valued functions of a vector. Namely that if $a = a(\boldsymbol{x})$ and $\boldsymbol{v} = \boldsymbol{v}(\boldsymbol{x})$, then $$\nabla_{\boldsymbol{x}}(a\boldsymbol{v})=a\nabla_{\boldsymbol{x}}\boldsymbol{v}+\boldsymbol{v}\otimes\nabla_{\boldsymbol{x}}a.$$ If you haven't seen this ...


2

Hint: From the quotient rule of differentiation $$\frac{d}{dx}\tan x = \frac{d}{dx}\frac{\sin x}{\cos x} = \frac{\cos x\cos x - \left(-\sin x \sin x\right)}{\cos^2 x} = \frac{1}{\cos^2 x}$$


2

$$\frac1{|x|}-\frac1{\sqrt{x^2+R^2}}=\frac{\sqrt{x^2+R^2}-|x|}{|x|\sqrt{x^2+R^2}}=\frac{x^2+R^2-x^2}{|x|\sqrt{x^2+R^2}(\sqrt{x^2+R^2}+|x|)}.$$ For $|x|\gg R$, $\sqrt{x^2+R^2}\approx|x|$ and the expression simplifies to $$\frac{R^2}{2|x|^3}.$$


2

Since \begin{eqnarray} \frac{\partial F}{\partial x}(0,0)&=&\frac{\partial f}{\partial x}(0,0)+2\frac{\partial f}{\partial y}(0,0)=4+2(-3)=-2\\ \frac{\partial F}{\partial y}(0,0)&=&3\frac{\partial f}{\partial x}(0,0)-\frac{\partial f}{\partial y}(0,0)=3(4)-(-3)=15, \end{eqnarray} we have \begin{eqnarray} \frac{\partial ...


1

$f'(c,u)=\nabla f(c) . u=(4,-3). (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})= 4\frac{\sqrt{2}}{2}-3\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}$ ( "." is inner product symbole.)


1

The width will be the circumference of the base circle, hence its radius will be $\dfrac{w}{2\pi}$ and its area $\dfrac{w^2}{4\pi}$, and the volume of the cylinder: $$V=\smash{\dfrac{w^2}{4\pi }} (19-w)$$ Now$$\frac{\mathrm d\mkern0.5mu V}{\mathrm d\mkern0.5mu w}=\smash{\dfrac1{4\pi}}\bigl(2w(19-w)-w^2\bigr)=\smash{\dfrac{w(38-3w)}{4\pi}}$$ Furthermore, the ...


1

Let $g(x)=\log\left(\frac{e^x+1}{e^{x}-1}\right)$: $g$ maps $\mathbb{R}^+$ into $\mathbb{R}^+$, $g(g(x))=x$ and the only fixed point of $g$ is $x_0=\log(1+\sqrt{2})$. For any $x\in\mathbb{R}^+\setminus\{x_0\}$, let $y=g(x)$. We have to fulfill: $$ f(x)^2\,f(y) = a x,\qquad f(x)\,f(y)^2 = a y $$ hence by dividing the first identity by the square root of the ...


1

There is no reason why you can't set the variables as you did, and why you should not get the right answer if you proceed properly ! By similar triangles, $\dfrac{y}{6} = \dfrac{x}{15-6}$ which yields $y = \dfrac{2x}{3}$ so, (b) $\dfrac{dy}{dt} = \dfrac23 \cdot \dfrac{dx}{dt} =\dfrac{10}{3}$ ft/s and (a), $\dfrac{dx}{dt} + \dfrac{dy}{dt} = \dfrac{25}{3}$ ...


1

For a monic polynomial $p(x)=x^n+a_{n-1}x^{n-1}+\ldots+a_1x+\color{red}{a_0}$ take the Frobenius companion matrix $$ A=\left[\matrix{0 & 0 & \ldots & 0 & \color{red}{-a_0}\\ 1 & 0 & \ldots & 0 & -a_1\\ 0 & 1 & \ldots & 0 & -a_2\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & ...


1

$$ \int\frac{1}{e^{2x}+e^x} \,dx = \int\frac{e^x\,dx}{e^{3x} + e^{2x}} = \int \frac{e^x\,dx}{(e^x)^2(e^x + 1)} = \int\frac{du}{u^2(u+1)} $$ Then use partial fractions.


1

Let $e^x= t\;,$ Then $\displaystyle e^x dx = dt\Rightarrow dx = \frac{1}{e^x}dt = \frac{dt}{t}$ So Integral $$\displaystyle I = \int\frac{1}{e^x(e^x+1)}dx = \int\frac{1}{t^2(t+1)}dt$$ Now Let $\displaystyle t = \frac{1}{u}\;,$ Then $\displaystyle dt = -\frac{1}{u^2}du$ So we get $$\displaystyle I = -\int \frac{u}{u+1}du = -\int\frac{(u+1)-1}{u+1}du = ...


1

The integral can be written: $$ \int^{\infty}_{0}\frac{x^n}{x^{m+n+1}}dx=\int^{\infty}_{0} x^nx^{-m-n-1} dx \\ =\int^{\infty}_{0} x^{-m-1} dx=\int^{\infty}_{0} \frac{1}{x^{m+1}} dx $$ which does not converge. You may be confusing the integral you have with the Beta function, which has the representation $$ ...


1

Yes. Every piecewise-smooth curve can be re-parametrized by length. Differentiate your equation and you will see that it is equivalent to the condition that $\|v'\| = 1$ for all $t$. A well-known example is the circle $(\cos t, \sin t)$.


1

A slightly different approach (in attachment) leads to : $$f(r)=\frac{1}{c_1}\ln \left| c_1r \pm \sqrt{c_1^2r^2-1}\right|+c_2$$ which includes all real solutions, not forgeting $f(r)=c_2$ at limit $c_1 \rightarrow \infty.$


1

L'Hospital's rule does not work here and we can directly work using the substitution $t = h^{2}$ as follows: \begin{align} L &= \lim_{h \to 0}\frac{f(x + 3h^{2}) - f(x - h^{2})}{2h^{2}}\notag\\ &= \lim_{t \to 0^{+}}\frac{f(x + 3t) - f(x - t)}{2t}\notag\\ &= \lim_{t \to 0^{+}}\frac{f(x + 3t) - f(x) + f(x) - f(x - t)}{2t}\notag\\ &= \lim_{t \to ...


1

The integral of $\dfrac1{\cos^2 x}$ is $\tan x$, not the function itself. After the O.P.'s edit: It is because the derivative of $\tan x$ is $\dfrac1{\cos^2x}$ (and it is also $1+\tan^2x$).


1

The textbook answer has a mistake. They probably wrote $f'(p)/f(p)=-2$ where it should be $f'(p)/f(p)=-{1\over 2}$. You can see that their answer is wrong because the tangent line to the graph of $f(x)=2e^{-x}$ at $x=0$ is $y=2-2x$, which intersects the $x$-axis at $x=1$, and not at $x=2$, as the assumption indicates.


1

(1) $$\int_{-\sqrt 2}^{-2} \frac{\operatorname d y}{y\sqrt{y^2-1}}$$ Substitute $s^2 = y^2-1$ and $s\operatorname d s = y\operatorname d y$. $$\int_{1}^{\sqrt 3}\frac{s\operatorname d s}{y\cdot y\sqrt{s^2}}$$ $$\int_{1}^{\sqrt 3}\frac{\operatorname d s}{s^2+1}$$ Use $\dfrac{\operatorname d \arctan s}{\operatorname d s} = \dfrac{1}{s^2+1}$ ...



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