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\begin{align} \int_0^1\frac{x^3-x^2}{\ln x }\mathrm{d}x &=\int_0^1\frac{x^3-1-x^2+1}{\ln x }\mathrm{d}x\tag{1}\\ &=\int_0^1\frac{x^\color{blue}{3}-1}{\ln x }\mathrm{d}x-\int_0^1\frac{x^\color{red}{2}-1}{\ln x}\mathrm{d}x\tag{2}\\ &=\ln(\color{blue}{3}+1)-\ln(\color{red}{2}+1)\tag{3}\\ &=\ln\frac43\tag{4}\\ \end{align} ...

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We want to figure out when $e^{-x} = x$. Multiplying by $e^x$ gives $$xe^x = 1$$ The solution to this equation is defined as the $\Omega$-constant, and shares many interesting properties. So we have $\Omega e^\Omega=1$ or $e^{\Omega} = 1/\Omega$ or $\Omega=e^{-\Omega}$. Several fast approximations can be found at the link above. The integral is then ...

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A typical way to prove that a function $f$ is constant is to show that $$\int_0^\pi f(x)g(x)\,dx=0\tag{1}$$ for every function $g\in C[0,\pi]$ with zero mean (i.e., $\int_0^\pi g(x)\,dx=0$). Indeed, if $f(x_1)\ne f(x_2)$, then let $g$ be the function that is zero on most of the interval and has two triangular peaks at $x_1,x_2$: one positive, one negative. ...

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This is not an answer, but I would like to share what I did on this. Let $s=1-w^2$ and $y=z^2$ to have $$I=4\int_0^1\int_0^1\frac{wz\arcsin wz}{\sqrt{1-z^2}(1-w^2z^2)}dwdz$$ Now using $wz=u$ and $z=v$ we obtain $$I=4\int_0^1\int_0^v\frac{u\arcsin u}{v\sqrt{1-v^2}(1-u^2)}dudv$$ Approach $1$: write both $\arcsin u$ and $(1-u^2)^{-1}$ in terms of Taylor's ...

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The change of variables should be in the form not $x=g(θ)$, but $\theta=g(x)$, so you should fix a branch $g$ of the multivalued function $\arccos$.

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The conversion between cylindrical and Cartesian coordinates is $x = r\cos\theta$, $y = r\sin\theta$, $z = z$. Thus, the density $\rho(x,y,z) = 3-z$ (Cartesian) becomes $\rho(r,\theta,z) = 3-z$ (cylindrical).

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Normally, just like $$\iiint_\Omega dV = \text{Volume}(\Omega),$$ so too $$\iiint_\Omega \rho(x,y,z) dxdydz = \text{Mass}(Omega),$$ where $\rho$ denotes density and $\Omega$ denotes your region.

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Separating your variables appropriately, we have $$\int_0^T \delta(t-t_j)e^{-(s+\mu\lambda^2)t}\,dt\int_0^l \delta(x-R)\varphi(x)\,dx.$$ Since $0 < R < l$, the sifting property of the Dirac delta gives us $\varphi(R)$ for the second integral. The first is very similar since $0 < t_j < T$.

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Here is a complex-analytic method: Notice that $$\int_{0}^{\infty} \log\left( \frac{x^{2} + 2x\cos b + 1}{x^{2} + 2x\cos a + 1} \right) \frac{dx}{x} = 2 \int_{0}^{\infty} \Re \frac{\log(1 + e^{ib}x) - \log(1 + e^{ia}x)}{x} \, dx. \tag{1}$$ Let $R$ be a positive large number. Then the function $z \mapsto \log(1+z)/z$ is analytic on $\Bbb{C} \setminus ... 5 Another approach is to use the Fourier series $$\sum_{k=1}^{\infty}\frac{x^{k} \cos(ka)}{k} = - \frac{1}{2} \log \left(x^{2} - 2 x \cos(a) +1 \right) \ , \ |x| <1$$ which can be derived from the Maclaurin series of$\log(1-z)$by replacing$z$with$xe^{ia}\$ and equating the real parts on both sides. \begin{align} ... 5 Let us first substitute y = x^{2} + 1. Then \begin{align*} I &:= \int_{0}^{1} \frac{x^{3}}{2(2-x^{2})(1+x^{2}) + 3\sqrt{(2-x^{2})(1+x^{2})}} \, dx \\ &= \frac{1}{2} \int_{1}^{2} \frac{y - 1}{2y(3-y) + 3\sqrt{y(3-y)}} \, dy \tag{1} \end{align*} Using the substitution y \mapsto 3-y, it follows that I = \frac{1}{2} \int_{1}^{2} ...

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