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There was a little mistake made when squaring, you should have $$\frac{y^6+2+y^{-6}}{4}\tag{1}$$ inside the square root. And the square root of (1) is very nice, in our interval it is $$\frac{y^3+y^{-3}}{2}.$$ Remark: Many arclength exercises, including this one, are rather contrived. The coefficients were carefully chosen to make the thing we are ...
Let $\displaystyle\;u(x) = \frac{x^2-13x-1}{x}\;$. As $x$ varies over $\mathbb{R}$, we have u(x) increases monotonically from $-\infty$ at $-\infty$ to $+\infty$ at $0^{-}$. u(x) increases monotonically from $-\infty$ at $0^{+}$ to $+\infty$ at $+\infty$. This means as $x$ varies, $u(x)$ covered $(-\infty,\infty)$ twice. Let $x_1(u) < 0$ and $x_2(u) ... 5 HINT: For$a$fixed$\int_{-\infty}^\infty\exp\left(-\frac{(x^2+sx-b)^2}{a x^2}\right)\ dx$is constant in$s$and$b\ge 0$.$\bf{Added:}$The function$\frac{x^2 + s x - b}{x} = x - \frac{b}{x} + s$invariates the Lebesgue measure as @achille hui: showed in his answer. Let$n \in \mathbb{N}\alpha_1$,$\ldots$,$\alpha_ndistinct real ... 4 Adding another solution owing to a friend of mine. Through some algebra, the integral is equivalent to $$\int_{-\infty}^\infty \exp\left(-\frac1{611}\left((x-x^{-1})-13\right)^2\right)\ dx$$ Then using the following identity $$\int_{-\infty}^\infty f(x-x^{-1})\ dx = \int_{-\infty}^\infty f(x)\ dx$$ We have \begin{align} &\int_{-\infty}^\infty ... 4\begin{align}\int_0^\infty\operatorname{Ei}^3(-x)\,dx&=-3\operatorname{Li}_2\left(\frac14\right)-6\ln^22.\\\\\int_0^\infty\operatorname{Ei}^4(-x)\,dx&=24\operatorname{Li}_3\left(\frac14\right)-48\operatorname{Li}_2\left(\frac13\right)\ln2-13\,\zeta(3)\\&-32\ln^32+48\ln^22\,\ln3-24\ln2\,\ln^23+6\pi^2\ln2.\end{align}$$4 The logarithmic series is related to the Alladi-Grinstead constant, and its decimal expansion can be found on OEIS. It has no known closed form, but it can also be expressed as ~\displaystyle\sum_{n=1}^\infty\frac{\zeta(n+1)-1}n. However, your integral is not equivalent to it, but rather to the similar ... 4 The ideal substitution in this case would be \sqrt{x^2+y^2}+y=t. The substitution relation gives t as a function of y, but we also need to know y as a function of t. We can isolate y as follows:$$\begin{align} \sqrt{x^2+y^2}+y&=t\\ \sqrt{x^2+y^2}&=t-y\\ x^2+y^2&=(t-y)^2\\ x^2+y^2&=t^2-2ty+y^2\\ x^2&=t^2-2ty\\ ... 3 After your first change of variable you get $$ac \int \dfrac{y}{\frac{cd}{b}-y} \dfrac{1}{(1-y)}dy$$ Now forgetting about the constants for a moment, $$\int \dfrac{y}{(A-y)(B-y)} dy = \dfrac{1}{A-B}\int \dfrac{B}{B-y} - \dfrac{A}{A-y} dy$$ Then go back and substitute. 3 Process 1: It is known that \begin{align} \int_{0}^{1} x^{\nu -1} (1+x)^{\lambda} (1-x)^{\mu -1} \, dx = B(\mu, \nu) \, {}_{2}F_{1}(- \lambda, \nu; \mu+\nu; -1) \end{align} for which \begin{align} \int_{0}^{1} \left[ \ln(x) \, \ln(1-x) \, \ln(1+x) \right]^{2} \, dx = \partial_{\nu}^{2} \partial_{\mu}^{2} \partial_{\lambda}^{2} \left[ B(\mu, \nu) \, ... 3 Your way is fine, it's just a matter of recognising a differentiated geometric series in that: \begin{align} \sum_{n=0}^\infty \frac{n(e-1)}{e^{n+1}} &= \sum_{n=0}^\infty \frac{e-1}{e^2} \cdot \frac{n}{e^{n-1}}\\ &= \frac{e-1}{e^2} \sum_{n=0}^\infty n\left(\frac{1}{e}\right)^{n-1}\\ &= \frac{e-1}{e^2} \frac{1}{\left(1-\frac{1}{e}\right)^2}\\ ... 3 Our integral is\int_0^\infty xe^{-x}\,dx-\int_0^\infty \left(x-\lfloor x\rfloor\right)e^{-x}\,dx.\tag{1}$$The first integral in (1) has value 1. For the second integral, split as you did as a sum of integrals$$\int_n^{n+1} (x-n)e^{-x}\,dx.$$Integrate. We get \frac{e-2}{e}e^{-n}. Summing from 0 to \infty we get \frac{e-2}{e-1}. Thus the ... 3 I(5) doesn't seem to have a closed form that I could find. On the other hand, I found numerically that$$ \begin{eqnarray}I(4) &=& -\tfrac{32}{3} \text{Li}_4(\tfrac{1}{2})+\tfrac{52}{3} \zeta(4)+\tfrac{8}{3} \zeta (2) \log^22-\tfrac{4}{9}\log^4 2. \end{eqnarray}$$Case 3. The way to do the integral I(3) is to write$$ \mathrm{li}(x) = ... 3 The given result is wrong, if the integral is interpreted as a principal value integral. It doesn't make sense for any natural interpretation of an integral that the value of a real integral should be a nonzero purely imaginary number anyway, so also for other interpretations, the given result is most likely wrong. One way to compute the integral (as a ... 3 As an alternate solution : why not to start using\cos(2x)=1-2\sin^2(x)$and$\sin(2x)=2 \sin(x)\cos(x)$. So, $$I=\int {\sin^2x \over 1 + \sin x\cos x}dx=\int \frac{1-\cos(2x)}{2+\sin(2x)}dx$$ Now, use the tangent half-angle substitution$t=\tan(x)$to get $$I=\int\frac{4 t^2}{t^4+t^3+2 t^2+t+1}dt$$ Using partial fraction decomposition $$\frac{4 ... 3 Generally. If f:[0,1]\to\mathbb{R} is a continuous function then$$ \int_0^{1/n}f(t)dt\sim \frac{f(0)}{n}. $$Indeed, if ~\displaystyle F(x)=\int_0^xf(t)dt, then it is easy to prove that F'(0)=f(0). we just write$$\left\vert\frac{F(x)}{x}-f(0)\right\vert\leq \int_0^1\left\vert f(tx)-f(0)\right\vert dt $$So, the the continuity of f at 0 proves ... 3 Too much work for something really trivial. An affine map sends the ellipsoid into the unit sphere, whose volume is well-known (\frac{4\pi}{3}). Hence you just have to compute the determinant of a diagonal Jacobian matrix (abc). Profit. 2 Basically, you make a change of variable t=\tan(x); doing so, you have$$\int _0^{\frac{\pi }{4}}\:\left(\frac{1}{\cos^4x-\cos^2x\sin^2x+\sin^4x}\right)\:dx=\int _0^1\frac{t^2+1}{t^4-t^2+1}dt$$and$$\int\frac{t^2+1}{t^4-t^2+1}dt=\tan ^{-1}\left(\frac{t}{1-t^2}\right)$$I must say that I do not see where the 3 disappeared. 2 Using \displaystyle\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx, \displaystyle I=\int_0^\frac\pi2\frac{\sin^2x}{1+\sin x\cos x}dx=\int_0^\frac\pi2\frac{\cos^2x}{1+\sin x\cos x}dx \displaystyle2I=\int_0^\frac\pi2\frac1{1+\sin x\cos x}dx=\int_0^\frac\pi2\frac{\sec^2x}{1+\tan^2x+\tan x}dx Setting \tan x=u ... 2 Using identity in @user137794's answer: $$\int_{-\infty}^\infty f(x-x^{-1})\ dx = \int_{-\infty}^\infty f(x)\ dx$$ where the complete proof can be seen here. The problem can be generalised to evaluate $$\int_{-\infty}^\infty \exp\left(-\frac{(x^2-bx-1)^2}{ax^2}\right)\ dx = \sqrt{a\pi}$$ Proof: ... 2 Set g(x)=\int_0^x f(t)\,dt. Then g(0)=0, and g(x)\ne 0, for all x>0, and as 2g(x)=\left(f(x)\right)^2\ge 0, then g(x)>0, for all x>0. Also, g is differentiable, as the indefinite integral of a continuous function, and so is its square root \sqrt{2}\,g^{1/2}(x)=f(x), for x>0. So g satisfies g'=\sqrt{2g}, for x>0, and ... 2$$\int_{e^{-1}}^e\dfrac{dt}{t(1+|\ln t|)^2}$$use the substitution$$x=\ln t.$$Then dx=\dfrac{dt}{t} and x \to 1,-1 as t\to e,e^{-1}. Now our integral become to$$\int_{-1}^1\dfrac{dx}{(1+|x|)^2}=\int_{-1}^0\dfrac{dx}{(1-x)^2}+\int_{0}^1\dfrac{dx}{(1+x)^2}.$$I think you can continue from here. 2 The integral is \int_{e^{-1}}^{e}\frac{dt}{t(1 + |\ln t|)^2} Sub t = e^x$$\int_{e^{-1}}^{e}\frac{dt}{t(1 + |\ln t|)^2} = \int_{-1}^{1} \frac{e^xdx}{e^x(1+|x|)^2} = \int_0^1\frac{dx}{(1+x)^2} + \int_{-1}^0 \frac{dx}{(1-x)^2} = -\frac{1}{2} + 1 + 1 - \frac{1}{2} = 1$$2 Let$$y=x^x\implies \ln y=x\ln x\\ y'=x^x\{\ln x+1\}\\ \int_1^e x^x\{\ln x+1\}dx=\{x^x\}_1^e$$2 Keep kicking in and out (and balancing) constants and applying your candidate formula:$$6\sqrt{2}\int_{0}^{\pi/6}\sqrt{\frac{1-\cos\,6x}{2}}\,dx = 6\sqrt{2}\int_{0}^{\pi/6}|\sin 3x|\,dx$$2 Another method consists in using a Fourier series : Of course, it's not so easy, but it's a nice exercise ! 2 In terms of multiple zeta values this integral is equal to$$ 2 \zeta (\bar3,\bar1)+\zeta (\bar3,1)+6 \zeta (\bar2,\bar1)+3 \zeta (\bar2,1)+24 \zeta (\bar1,\bar1)+12 \zeta (\bar1,1)-6 \zeta (2,\bar1)-2 \zeta (3,\bar1)+\zeta(\bar3,\bar1,\bar1)+\zeta (\bar3,\bar1,1)+\zeta (\bar3,1,\bar1)+2 \zeta (\bar2,\bar1,\bar1)+2 \zeta (\bar2,\bar1,1)+2 \zeta ... 2 $$\frac1{25e^x+9}=\frac{e^{-x}}{25+9e^{-x}}$$ Set$25+9e^{-x}=u$Alternatively, setting$25e^x+9=v,25e^x\ dx=dv\iff dx=\dfrac{dv}{v-9}$$$\implies I=\int\frac{dx}{25e^x+9}=\int\frac{dv}{v(v-9)}$$ $$\implies 9I=\int\frac{v-(v-9)}{v(v-9)}dv=\cdots$$ 2 Hint $$\int \frac{\log(1+x)}{x} \, \mathrm dx=-\text{Li}_2(-x)$$ Added later Using the trick given in the hint, then $$\int \frac{\operatorname{Li}_2(-x)}{x} \log(1+x)\, \mathrm dx=-\frac{\text{Li}_2(-x){}^2}{2}$$ and so $$\int^1_0 \frac{\operatorname{Li}_2(-x)}{x} \log(1+x)\, \mathrm dx=-\frac{\pi ^4}{288}$$ 2 Contour integration is a bit less painful. For first, it is better to write our integral as: $$I = \frac{1}{4}\int_{0}^{2\pi}\frac{\cos(3x)+\cos(5x)}{(2-\cos x)^2}\,dx,$$ then, since$\cos x = \frac{e^{ix}+e^{-ix}}{2}$, by setting$z=e^{ix}$we get: $$I = -\frac{i}{4}\left(\oint\frac{2(z^6+1)}{z^2(z^2-4z+1)^2}\,dz + ... 1 The problem is that$$dx=t\sqrt{t^2+4}dt$$and you have to integrate w.r.t$x\$