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Set $t=\frac{e^x}{4}$ , or $e^x=4t$ , thus $4dt=e^x dx$ as $x\to-\infty$ then $t\to 0$ as $x\to\ln(4)$ then $t\to 1$ we have $$I=\int_{-\infty}^{\ln(4)}\frac{xe^x}{\sqrt{4e^x-e^{2x}}}dx=\int_{0}^{1}\frac{\ln(4t)}{\sqrt{16t-16t^2}}4dt=\int_{0}^{1}\frac{\ln(4)+\ln(t)}{\sqrt{t-t^2}}dt$$ $$I=\ln(4)\int_{0}^{1}\frac{1}{\sqrt{t-t^2}}dt+\int_{0}^{1}\frac{\ln(t)... 7 It's good, if read correctly; you are less likely to make errors with this if you set$$ G(x)=\int_0^xg(t)\,dt $$and write$$ \int_0^af(x)g(x)\,dx= \Bigl[f(x)G(x)\Bigr]_0^a-\int_0^a f'(x)G(x)\,dx $$If you prefer not to use G, you can write$$ \int_0^af(x)g(x)\,dx= \left[f(x)\int_0^x g(t)\,dt\right] - \int_0^a f'(x)\left(\int_0^x g(t)\,dt\right)\,dx $$5 We can use an integral representation of the Dirichlet eta function to show that$$\int_{0}^{\infty} \frac{\tanh^{2}(x)}{x^{2}} \, dx = \int_{0}^{\infty} \left(1-\frac{1}{\cosh^{2}(x)} \right) \frac{dx}{x^{2}} = -56 \, \zeta'(-2) = \frac{14 \, \zeta(3)}{\pi^{2}}. $$An integral representation of the Dirichlet eta function is$$\eta(s) = \frac{1}{\Gamma(s)...

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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \... 5 We can write the interal in$(4)as $$I=-\int_{0}^{1}\frac{1-x+\log\left(x\right)}{\log^{2}\left(x\right)}dx$$ now define $$I\left(\alpha\right)=-\int_{0}^{1}\frac{x^{\alpha}\left(1-x+\log\left(x\right)\right)}{\log^{2}\left(x\right)}dx,\,\alpha\geq0 .$$ We have $$I''\left(\alpha\right)=-\int_{0}^{1}x^{\alpha}\left(1-x+\log\left(x\right)\right)dx=-\... 5 Through the substitution x=e^{-t} the original integral equals:$$ I=\int_{0}^{+\infty}\left(2\frac{e^{-t}-1}{t^2}+\frac{e^{-t}+1}{t}\right)e^{-t}\,dt \\=2\color{purple}{\int_{0}^{+\infty}\frac{e^{-t}-1+t}{t^2}\,e^{-t}\,dt}+\color{blue}{\int_{0}^{+\infty}\frac{e^{-t}-1}{t}\,e^{-t}\,dt}$$where the blue integral is yet manageable through Frullani's theorem (... 5 Hint. One may observe that$$ (x^2+3x+2)=(x+1)(x+2) $$leading to the following partial fraction decomposition$$ \begin{align} \frac{ax+b}{(x^2+3x+2)^2}&=\frac{ax+b}{(x+1)^2(x+2)^2} \\\\&=\frac{-a+b}{(x+1)^2}+\frac{3 a-2 b}{x+1}+\frac{-2 a+b}{(x+2)^2}+\frac{-3 a+2 b}{x+2}. \end{align} $$Integrating each term gives$$ \int_0^1\frac{ax+b}{(x^2+3x+2)^... 5 Hint: Try the subx \mapsto \frac{\pi}{2} - x$to get $$\int_0^{\pi/2} \frac{(\pi/2 - x) \cot x}{\csc x + \cot x} \, \mathrm{d}x = \int_0^{\pi/2} \frac{\pi/2 - x}{1 + \sec x} \, \mathrm{d}x$$ Then the rest is do-able using $$\frac{1}{\sec x + 1} = \frac{\cos x}{\cos x + 1} = \frac{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}{2\cos^2 \frac{x}{2}} = \frac{1}{2} -... 4 This question, or one very similar to it, appeared on a diagnostic examination for incoming freshmen at the California Institute of Technology. The first observation is that the system is underdetermined: for we may factor out b and let r = a/b to express the integrand as$$f(x) = b \cdot \frac{rx+1}{(x^2+3x+2)^2}.$$For r > 0 and x \in [0,1], ... 3 Generally,$$ \int_0^af^\prime(x)\,dx=f(a)-f(0) $$so if$$ f(a)=\int_0^a g(x)f(x)\,dx $$and f(0)=0 and furthermore if it is generally true that$$ f(x)=\int_0^xg(t)f(t)\,dt$$then one might conclude that$$f^\prime(x)=g(x)f(x)$$which implies$$ \dfrac{f^\prime(x)}{f(x)}=g(x)$$or$$ \left(\ln(f(x)\right)^\prime=g(x)$$so that$$f(x)=\exp\... 3 Here is a slightly different variation to OPs example, followed by another one. Suppose$p$is an even function, i.e.$p(x)=p(-x)$and$q(x)q(-x)=1. Then \begin{align*} \int_{-a}^{a}\frac{p(x)}{1+q(x)}\,dx=\int_{0}^{a}p(x)\,dx \end{align*} A proof of the statement together with an application can be found in this answer. Note: This technique ... 3 This really isn't so bad. $$\int_0^i\frac{i}{(i+x^2)^{3/2}}dx=\frac{i}{\sqrt{i(i+1)}}.$$ So you're summing: $$\sum_{i=m}^n\frac{i}{i+1}.$$ The latter sum unfortunately doesn't have an explicit form, unless you're willing to use digamma functions. 3 Consider the integral $$I= - \int_0^1 \frac{\ln(1-x^2)}{\sqrt{1-x^2}} (\sin^{-1} x)^4 \,dx.$$ Since $$(\sin^{-1} x)^4 = \frac32 \sum_{n=1}^{\infty} \cfrac{2^{2n} H_{n-1}^{(2)}}{n^2 \binom{2n}{n}} \,x^{2 n} \tag{1}$$ and $$-\int_0^1 \frac{\ln(1-x^2)}{\sqrt{1-x^2}} x^{2n}\,dx= \frac{\pi}{2} \binom{2n}{n} \frac{(H_n + 2\ln2)}{2^{2n}}, \tag{2}$$ we have $$\... 2 Yes!It is correct Let u=f(x), dv=g(x)dx, du=f'(x)dx, v=\int_{0}^{a}g(x)dx so \int_{0}^{a}f(x)g(x)dx=[f(x)\int_{0}^{x}g(x)dx]_{0}^{a}-\int_{0}^{a}[\int_{0}^{x}g(x)dx]f'(x)dx Next, integrate by parts a second time, u=f'(x)dx, dv=\int_{0}^{a}g(x)dx du=f''(x)dx, v=\int_{0}^{a}\int_{0}^{a}g(x)(dx)^{2} \int_{0}^{a}f(x)g(x)dx=[f(x)\int_{0}^{x}g(... 2 \newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \... 2 Given the formula for the area in polar coordinates and the symmetry of the configuration, the area you want to compute is just:$$ 2\pi + \int_{\pi/2}^{\pi}\left[2\left(1+\cos\theta\right)\right]^2\,d\theta =\color{red}{5\pi-8}.$$2 Because i is a constant with respect to the variable of integration, we can use the fact that for any constant b:$$\int \frac{b}{(b+x^2)^{3/2}}dx = \frac{x}{\sqrt{x^2 + b}}$$so that$$\int_0^i \frac{i}{(i+x^2)^{3/2}dx} = \frac{i}{\sqrt{i^2 + i}} = \sqrt{\frac{i^2}{i^2+1}}.$$When we square this (to get the ith term in the sum), we get:$$\left(\... 2 Any functiong(x)$such that$g(c+a)+g(c-a)=k$for all$a$on the interval$(0,b)$with any function$f(x)$such that$f(c-a)=f(c+a)$for all$a$on the interval$(0,b)will satisfy the equation $$\int_{c-b}^{c+b}{f(x)g(x)dx}=k*\int_{c}^{c+b}{f(x)dx}$$ because, using a trapezoidal Riemann sum after splitting the integrals into $$\int_{c-b}^{c}{f(x)g(x)dx}... 2 It appears that the integral when n=2 can be represented in terms of elliptic integrals:$$ I(2)=\frac{\pi}{2}-\frac{1}{\sqrt{6}}\left(\Pi\left(\frac23\mid\frac13\right)-K\left(\frac13\right)\right). $$Here the arguments of elliptic functions follow Mathematica conventions: that is,$$ K(m)=\int^{\pi/2}_{0}\frac{d\theta}{\sqrt{1-m\sin^2\theta}} and ... 2 Let the sum be y(s) and F'(s)=f(s). Assume converging, we have \begin{align*} y(s) &= \int_{0}^{s} f(t)[1+ay(t)]\, dt \\ y'(s) &= f(s)[1+ay(s)] \\ y'(s)-af(s)y(s) &= f(s) \\ I(s) &= \exp \left[-\int a f(s)ds \right] \quad \quad \text{(integrating factor)} \\ &= e^{-a F(s)} \\ I(s) y(s) &= \int f(s) I(s) ds \\ e^... 2\int \frac{dr}{r^2} \frac{1}{\sqrt{-(\frac{1}{r} - \frac{1}{p})^2 + \frac{\epsilon^2}{p^2} }}=-\int \frac { d\left( \frac { 1 }{ r } -\frac { 1 }{ p } \right) }{ \sqrt { \frac { \epsilon ^{ 2 } }{ p^{ 2 } } -\left( \frac { 1 }{ r } -\frac { 1 }{ p } \right) ^{ 2 } } } =-\frac { \epsilon }{ p } \int { \frac { d\frac { p\left( \frac { 1 }{ r } -\frac { ... 2 Hint...substitute $$\frac 1r-\frac 1p=u\cdot \frac {\epsilon}{p}$$ and then you have a standard\arcsin-type integral 2 You correctly pointed out the problem if the binomial expansion is used. If fact, you need to use hypergeometric functions since $$\int\left(1-\frac{1}{(1+x)^M}\right)x^{-\frac{2}{\alpha}-1}\,dx=\frac{1}{2} \alpha x^{-2/\alpha} \left(\, _2F_1\left(-\frac{2}{\alpha},M;1-\frac{2}{\alpha};-x\right)-1\right)$$ Using a CAS, what was obtained is \int_0^\... 2 There might be an easier solution exploiting symmetry but the one follows is what I came up: \begin{align*} \int_{0}^{\pi/2} \frac{x\tan x}{\sec x + \tan x} \, {\rm d}x &= \int_{0}^{\pi/2} \frac{x \tan x}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}} \, {\rm d}x \\ &= \int_{0}^{\pi/2} \frac{x \cos x \frac{\sin x}{\cos x}}{1+\sin x} \, {\rm d}x\\ &... 2 In the absence of any tricks, it looks feasible to do this integral through integration by parts to eliminate the x. I'm guessing there's a typo in the question. Assuming the integral portion is correct, we multiply and divide by \sec x-\tan x.\int_0^{\frac\pi2}x(\sec x\tan x-\tan^2x)dx=\int_0^\frac{\pi}2x(\sec x\tan x-\sec^2x+1)dx=x(\sec x-\... 1 Use Weierstrass substitution By settingt= \tan \left( \frac x2 \right) $The integral is equivalent to: $$2\int \frac{(1+t^2)^2}{(3-t^2)^3} dt$$ Decompose the fraction into: $$2 \int \left( \frac 1{3-t^2} - \frac 8{(3-t^2)^2} + \frac{16}{(3-t^2)^3} \right)dt$$ Since$\frac 1{3-t^2} = \frac 1{2\sqrt 3} \left( \frac 1{\sqrt 3 - t} + \frac 1{\sqrt 3 +...

1

On the first line you get $$v(t)=\frac{a_0}{\omega}\cdot (-\cos(\omega \cdot t))+C$$ then by putting $t=0$, you have $v(0)=0$ giving \begin{align} &0=\frac{a_0}{\omega}\cdot (-\cos(\omega \cdot 0))+C \\\\&0=-\frac{a_0}{\omega}+C \\\\&C=\frac{a_0}{\omega}. \end{align} Then there is no more contradiction with your second computation.

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As it is currently written, this function is a constant. What you would really need for this to be a function is something like this: $$f(x)=\int_a^xy(t)^{2}+2(\frac{dy}{dt})^{2}+(\frac{d^2y}{{dt}^2})^{2}dt$$ where $t$ is a dummy variable. Then, you can use the Fundamental Theorem of Calculus to find: $$\frac{df}{dx}=\lim_{t \to x}y(t)^{2}+2(\frac{dy}{dt})^{... 1 The total derivative of f\left(x(t),y(t)\right) is :$$\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt} Considering now the same function but depending on a parameter $a$, say $f_a\left(x(t),y(t)\right)$ : If $a$ isn't function of $t$ and if $t$ is the only variable the total derivative is the same : $... 1$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \...

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