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We can write $\displaystyle y = \frac{|x-3| + |x+1|}{|x+3| + |x-1|} = \frac{-x+3-x-1}{x+3-x+1} = -x+1\;,$ When $-3\leq x<-1$ Similarly $\displaystyle y = \frac{|x-3| + |x+1|}{|x+3| + |x-1|} =\frac{-x+3+x+1}{x+3-x+1} = 2\;,$ When $-1\leq x<1$ Similarly $\displaystyle y = \frac{|x-3| + |x+1|}{|x+3| + |x-1|} = \frac{-x+3+x+1}{x+3+x-1} = ... 3 HINT: Set$x^2=y$$$\int\dfrac{x^3}{1+x^6}dx=\int\dfrac y{1+y^3}\cdot\dfrac{dy}2$$ Now use partial fraction decomposition $$\dfrac y{1+y^3}=\dfrac A{1+y}+\dfrac{By+C}{1-y+y^2}$$ 0 Even though hyperbolic method is a good one, but I think there is one even simpler method, just take x^2 out of the radical then substitute (4-9/(x^2))=(t^2). 1 Let $$I = \int\frac{dx}{\sqrt{(4x^2-9)^3}}$$ and make the substitution$x = \frac{3}{2} \, \operatorname{sec}\thetato obtain \begin{align} I &= \frac{3}{2} \, \int \frac{\operatorname{sec}\theta \, \tan\theta \, d\theta}{ 27 \, \tan^{3}\theta} \\ &= \frac{1}{18} \, \int \frac{\operatorname{sec}\theta}{\tan^{2}\theta} \, d\theta = \frac{1}{18} \, ... 0 HINT...hyperbolic functions make this job easier. try substituting2x=3\cosh\theta$1 Hint: Expand the$\tan\theta$and$\sec\theta$to their respective$\sin\theta$,$\cos\theta$parts and cancel. You are left with an integral that can be solved using a$u-substitution. 1 Notice \begin{align} \frac{1}{\prod\limits_{j=0}^{k}(i+j)} = \frac{1}{k}\left(\frac{(i+k)-i}{\prod\limits_{j=0}^{k}(i+j)}\right) &= \frac{1}{k}\left[\frac{1}{\prod\limits_{j=0}^{k-1}(i+j)}-\frac{1}{\prod\limits_{j=1}^{k}(i+j)}\right]\\ &= ... 0 Compared to the numeric solution this is the correct solution: \iiint\limits_{\rm I\!R^3} \left( \sum_{i=0}^n a_i e^{-\frac{(x-b_i)^2+(y-c_i)^2+(z-d_i)^2}{2}} \right)^2 = \frac{1}{8}\pi^{\frac{3}{2}} \sum_{i=0}^n \left( a_i^2 erf(x-b_i) erf(y-c_i) erf(z-d_i) \right) + \\ \frac{1}{4}\pi^\frac{3}{2} \sum_{i=0}^n \sum_{j=i}^n \left( a_i a_j erf(x - ... 0 The Q of whether a function is Lebesgue-measurable or Lebesgue-integrable is unaffected by what values the function takes on any one set that has Lebesgue measure zero.Every countable (meaning finite or countably infinite) set of reals is Lebesgue-null (meaning that its Lebesgue measure is zero.) The set of rationals is countable, hence it's ... 0 As3-2x-x^2=2^2-(x+1)^2$$start with x+2=2\sin y where -\dfrac\pi2\le y\le\dfrac\pi2 \implies dx=2\cos y\ dy$$\int\dfrac{x\ dx}{\sqrt{3-2x-x^2}}=\int\dfrac{2(\sin y-1)2\cos y}{+2\cos y}\ dy=?$$5 OK, I believe there is. Here is my result:$$\displaystyle \sum_{i=1}^n \frac1{\displaystyle \prod_{j=0}^k (i+j)} = \frac1{k \cdot k!} \left [ 1 - \frac1{\displaystyle\binom{n+k}{k}} \right ] = \frac1{k} \left [\frac1{k!} - \frac1{(n+1)(n+2)\cdots(n+k)}\right ]$$This result checks out for all of the values of k and n I have plugged into ... 0 Take (e_i)_{i \in \mathbb{N}} enumeration of the rationals in [0,1]. Define a sequence f_n: [0,1] \rightarrow \mathbb{R} as follows: f_n(x)=1 if x \in \{e_1,...,e_n \} and f_n(x)=0 otherwise. Each f_n is continuous except at a finite number of points, hence riemann integrable. But the pointwise limit function is not riemann integrable, ... 1 Each \ln_a^b f_n is simply a number; so if each number is defined but the sequence of numbers doesn't converge, then extending from Riemann integrals to Lebesgue integrals (which will all have the same value) doesn't change anything. If you want an example where \int_a^b (\lim f_n) doesn't exist as a Riemann integral but does exist as a Lebesgue ... -1 Let \begin{eqnarray} U^+&=&\{(x,y) \in \mathbb{R}^2:\, f(x,y)> 0\}=\{(x,y)\in \mathbb{R}^2:\, x> 0\}\\ U^-&=&\{(x,y) \in \mathbb{R}^2:\, f(x,y)< 0\}=\{(x,y)\in \mathbb{R}^2:\, x< 0\}, \end{eqnarray} and consider the function f^\pm=|f|\mathbb{1}_{U^\pm}. Using polar coordinates x=r\cos\theta, y=r\sin\theta, we have ... 4 f is integrable if and only if 1<\alpha < 2. Polar coordinates are actually useful. Write (x,y)=r(\sin\theta,\cos\theta) and then you have$$f(r,\theta)=\sin\theta r^{1-2\alpha} \log(1+r^2)$$f is clearly continuous away from the origin so we need only check what happen near zero and "near" infinity. In a punctured neighborhood of the origin, ... 0$$\int\dfrac{1}{\cos^2{x}}dx =\int \dfrac{\cos^2{x}+\sin^2{x}}{\cos^2{x}}dx $$Letting u = \sin x \implies du = \cos x dx and v = \cos{x} \implies dv = -\sin{x} dx$$\int \dfrac{\cos^2{x}+\sin^2{x}}{\cos^2{x}}dx = \int \dfrac{vdu-udv}{v^2} = \int d \left(\dfrac{u}{v}\right) = \dfrac{u}{v} + C = \dfrac{\sin x}{\cos x}+C = \tan{x}+C$$1 Let \varepsilon > 0. Since \lim\limits_{x\to \infty} f(x) = a, there exists M > 0 such that |f(x) - a| < \varepsilon for all x \ge M. So for x > M,$$\left|\frac{1}{x}\int_0^x f(t)\, dt - a\right| \le \frac{1}{x}\int_0^M |f(t) - a|\, dt + \frac{1}{x}\int_M^x |f(t) - a|\, dt < \frac{C}{x} + \frac{\epsilon(x - M)}{x},$$where C = ... 0 METHOD 1: L'Hospital's Rule comes to the rescue.$$\begin{align} \lim_{x\to \infty}\frac{\int_0^x f(t)\,dt}{x}&=\lim_{x\to \infty}\frac{d}{dx}\int_0^x f(t)\,dt\\\\ &=\lim_{x\to \infty}f(x)=a \end{align}$$METHOD 2: If one does not wish to appeal to L'Hospital's Rule, then another approach is to take a similar way forward to that of the OP. ... 3 And yet, one more approach is$$\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}$$2 Hint: From the quotient rule of differentiation$$\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}$$1 The integral of \dfrac1{\cos^2 x} is \tan x, not the function itself. After the O.P.'s edit: It is because the derivative of \tan x is \dfrac1{\cos^2x} (and it is also 1+\tan^2x). 4 If a\in\mathbb{N}, factor 1+t^a then exploit:$$ \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 ... 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 Lete^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}duSo we get $$\displaystyle I = -\int \frac{u}{u+1}du = -\int\frac{(u+1)-1}{u+1}du = ... 3 Write$$\dfrac1{e^x(e^x+1)}=\dfrac{(e^x+1)-e^x}{e^x(e^x+1)}=e^{-x}-\dfrac1{e^x+1}$$and use How do I solve \displaystyle\int \frac{\mathrm{d}x}{e^x + 1} ? 5 HINT: Set e^x=u, x=\ln u\implies dx=\dfrac{du}u$$\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}$$3 Ok, i will give it a shot: Writing \int_{0}^{\infty}e^{-t(x+2)}=\frac{1}{x+2} and using \Im(e^{ix})=\sin(x) we may reformulate the problem as follows:$$ I=\Im\left[\int_0^{\infty}dte^{-2 t}\underbrace{\int_0^{\infty}dxe^{i\pi x^2-tx}}_{J(t)}\right] $$the inner intgral J(t) is quite straightforward (and also well known because it is just the laplace ... 1 The \Gamma function enters in the following way. Let f_a(x)=x^{a-1}e^{-x} for x>0 and 0 for x<0, with a>1. Then the Fourier transform is$$ \hat{f}_a(\xi)\;=\; \int_0^\infty x^{a-1}e^{-x} \,e^{-2\pi i x\xi}\;dx \int_0^\infty x^ae^{-x(1+2\pi i\xi)}\;{dx\over x} \;=\; (1+2\pi i\xi)^{-a}\Gamma(a) by the analytically continued form of the ... 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 ... 1 Assume k<0, otherwise the integral diverges for infinite T. We use a linear change of variable to normalize the exponent, with(f-ct)x+kx^2=k(x+\frac{f-ct}{2k})^2-\frac{(f-ct)^2}{4k}=-\left(\sqrt{|k|}x+\frac{f-ct}{2\sqrt{|k|}}\right)^2-\frac{(f-ct)^2}{4k}.Then ... 0 Letting \simeq denote the equality sign up to a constant, we have \begin{align*} \int_{x} \frac{e^{x}-1}{e^{x}+1} &= \int_{x}\frac{D (e^{x}+1)}{e^{x}+1} - \int_{x}\frac{1}{e^{x}+1}\\ &\simeq \log (e^{x}+1) - \int_{u := e^{x}} \frac{1}{(u+1)u}\\ &\simeq \log (e^{x}+1) - x + \log (e^{x}+1) = 2\log (e^{x}+1) - x. \end{align*} To check that the ... 0 First, write the integrand as \displaystyle\int\dfrac{e^x-1}{e^x+1}\,dx = \int\dfrac{(e^{x/2}-e^{-x/2})e^{x/2}}{(e^{x/2}+e^{-x/2})e^{x/2}}\,dx = \int\dfrac{e^{x/2}-e^{-x/2}}{e^{x/2}+e^{-x/2}}\,dx. You might notice that the numerator is almost like the derivative of the denominator, except missing a \tfrac{1}{2} factor. So substitute u = ... 1 Let\displaystyle I = \int\frac{e^x-1}{e^x+1}dx\;,$$Now let x=2t\;, Then dx = 2dt So Integral$$\displaystyle I = 2\int\frac{e^{2t}-1}{e^{2t}+1}dy = 2\int\frac{e^{t}-e^{-t}}{e^{t}+e^{-t}}dt\;,$$Now Put e^{t}+e^{-t} = u\;, Then (e^{t}-e^{-t})dt = du So$$\displaystyle I = 2\int\frac{1}{u}du = 2\ln|u|+\mathcal{C} = 2\ln ... 1 Here's a start: $$\frac{e^x-1}{e^x+1}=\frac{e^x+1-2}{e^x+1}=\frac{e^x+1}{e^x+1}+\frac{-2}{e^x+1}=1+\frac{-2}{e^x+1}.$$ Now use the substitutione^x+1\leadsto u$. 2 $$\int\dfrac{e^x-1}{e^x+1}dx=\int\dfrac{e^x+1-2}{e^x+1}dx$$ $$=\int dx-2\int\dfrac{dx}{e^x+1}$$ Method$\#1:$Let$e^x+1=v\implies e^x\ dx=dv\iff dx=\dfrac{dv}{v-1}$$$\int\dfrac{dx}{e^x+1}=\int\dfrac{dv}{v(v-1)}=\int\dfrac{v-(v-1)}{v(v-1)}dv=?$$ Method$\#2:$$$\int\dfrac{dx}{e^x+1}=\int\dfrac{e^{-x}\ dx}{e^{-x}+1}$$ Set$e^{-x}=y$2 The similarity between the sum$\displaystyle \sum_{k=-\infty}^{\infty}\arctan\left(\dfrac{1}{2k^{2}}\right)$and the integral$\displaystyle \int_{x=-\infty}^{\infty}\arctan\left(\dfrac{1}{2x^{2}}\right)\, dx$can be explained by Poisson's summation formula $$\sum_{k=-\infty}^{\infty}f(k) = \sum_{n=-\infty}^{\infty}\hat{f}(2\pi n)$$ where$fis a ... 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 $$... 1 The textbook answer has a mistake. They probably wrote f'(p)/f(p)=-2 where it should be f'(p)/f(p)=-{1\over 2}. You can see that their answer is wrong because the tangent line to the graph of f(x)=2e^{-x} at x=0 is y=2-2x, which intersects the x-axis at x=1, and not at x=2, as the assumption indicates. 2 Given the integral$$I(t) = \int_0^{\infty} e^{\frac{-t(s-1)^2}{2}} \left( \frac{t(s-1)^3}{3} \right) ds$$then by integration by parts$$I = \left[ -\frac{(s-1)^2}{3}e^{-t(s-1)^2/2}\right]_0^{\infty} + \frac{2}{3}\int_0^{\infty}(s-1)e^{-t(s-1)^2/2} ds.The remaining integral is t^{-1} times the derivative of the exponential and leads to \begin{align} I ... 0 I = \int sin^3 (2x) cos^2(2x) dx $$Substituting$$ t = 2x \\dt = 2 dx \\dx = \frac{dt}{2} $$So,$$ I = \frac{1}{2}\int sin^3(t) cos^2(t) dt $$I'm now going to factor out$$ sin^3 t = sin(t) (1 - cos^2 t) $$Because the factor of sin(t) will simplify the integral.$$ I = \frac{1}{2}\int sin(t)\; (1 - cos^2 (t)) \;cos^2(t) dt \\I = ... 2 HINT: Write\sin^3(2x)\cos^2(2x)=(1-\cos^2(2x))\cos^2(2x)\sin(2x)$SPOLIER ALERT: SCROLL OVER THE SHADED AREA TO REVEAL SOLUTION 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 ... 4 It's easy to check the correctness of an antiderivative by differentiating: $$\dfrac{d}{dx} \dfrac{1}{4} \text{arcsec}(x^2/2) = \dfrac{1}{x^3 \sqrt{1 - 4/x^4}} = \dfrac{1}{x \sqrt{x^4 - 4}}$$ So$\dfrac{1}{4} \text{arcsec}(x^2/2)+C$is indeed an antiderivative of$1/(x \sqrt{x^4-4}). I might note for completeness, however, that the domain of both ... 1 Consider \begin{align} I &= \int (5 - 5 \sin x) \cdot (5 + 5 \sin x) \, dx \\ &= 25 \, \int (1-\sin x) \cdot (1 + \sin x) \, dx = 25 \, J \end{align} Letdv = 1-\sin x$then$v = x + \cos x$,$u=1+\sin x$,$du = \cos xand \begin{align} J &= (x+\cos x)(1 + \sin x) - \int (x + \cos x) \, \cos x \, dx \\ &= (x + \cos x)(1 + \sin x) - \int x ... 3 Let $$\displaystyle I = \int \frac{1}{x\sqrt{x^4-4}}dx\;,$$ Letx^2=2\sec \phi\;,$Then$xdx = \sec \phi\cdot \tan \phi d\phi$So Integral $$\displaystyle I = \int\frac{x}{x^2\sqrt{x^4-4}}dx = \int\frac{\sec\phi\cdot \tan \phi}{2\sec \phi\sqrt{4(\sec^2 \phi-1)}}d\phi$$ Using the formula$\bullet\; \sec^2 \phi = 1+\tan^2 \phi\Rightarrow \sec^2 \phi = ...

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Let $$\displaystyle I = \int (5-5\sin x)\cdot (5+5\sin x)dx = 25\int (1-\sin^2 x)dx = 25\int \cos^2 xdx$$ Now Using Integration by parts, So $$\displaystyle I = 25\int \cos x\cdot \cos xdx = 25\cos x\cdot \sin x+25\int \sin x\cdot \sin xdx$$ Using $$\bullet\; \sin 2x = 2\sin x\cdot \cos x$$ So \displaystyle I = \frac{25}{2}\sin 2x+25\int (1-\cos^2 x)dx ... 0 However if you really want integration by parts: \begin{align} \int \sin(x) \cdot \sin(x) dx &= \sin(x) \cdot (-\cos(x))-\int \cos(x) \cdot (-\cos(x) )dx+C \\ &= -\sin(x) \cdot \cos(x)+\int \cos^2(x) dx +C \\ &= -\sin(x) \cdot \cos(x)+\int(1-\sin^2(x) )dx +C \\ &= -\sin(x) \cos(x)+x -\int \sin^2(x) dx +C. \end{align} And so add\int ...

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HINT: If $C(x)=x^2$, then we can find a closed form for its expected value. Recall that $$I(a)=\int_{-\infty}^{\infty}e^{-ax^2}dx=\sqrt{\frac{\pi}{a}}$$ and $$-I'(a)=\int_{-\infty}^{\infty}x^2e^{-ax^2}dx=\frac{\pi^{1/2}}{2a^{3/2}}$$ Then, change variables from $(x-\mu)/\sqrt{2\sigma}$ to $x$ and expand the quadratic inside the integral. Can you ...

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Here is an approach. We give a preliminary result. A series of squares of logarithms Let us consider the poly-Hurwitz zeta function initially defined by the series \begin{align} \displaystyle \zeta(s,t\mid a,b) := \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)^t}, \quad \Re a>-1, \, \Re b>-1, \, \Re (s+t)>1. \tag1 \end{align} ...

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