The Rubin Report: Liberal or Not?

I'm writing this article actually rather late in comparison to the story, but oh well! So WHAT IS THE STORY ... Dave Rubin and his show the Rubin Report are

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Integral from 0 to ∞ [x^a/(x^b+1)] dx

\begin{align*}& I(a, b) = \int_0^\infty \dfrac{x^a}{x^b+1} dx =\dfrac{\beta(\dfrac{a+1}{b}, 1-\dfrac{a+1}{b})}{b} \\& \\& {\rm{Proof}}: \\& \\& \therefore u=x^b \Longrightarrow \int_0^\infty \dfrac{u^{\frac{a-b+1}{b}}}{b(u+1)} du \\& \\& \therefore \int_0^\infty \dfrac{u^{\frac{a-b+1}{b}}}{b(u+1)} du = \dfrac{\beta(\dfrac{a+1}{b}, 1-\dfrac{a+1}{b})}{b} \\&

Integrals: Royal Properties

\begin{align*}& {\rm{Given}}:\int_0^\infty f(x) dx \\& \\& (1) \: \rm{Queen \: Property} \\& \\&\therefore u=(x)^{-1} \Longrightarrow\int_0^\infty f[(x)^{-1}] (x)^{-2} dx \\& \\& (2) \: \rm{Jacksonian \: Property} \\& \\&\therefore u=(x)^{-n} \Longrightarrown \int_0^\infty

Integral of [1+(x^g)]^(-n) from 0 to ∞

\begin{align*}& \int_0^\infty \dfrac{1}{(1+x^g)^n} dx =\int_0^\infty \dfrac{(x)^{gn-2}}{(1+x^g)^n} dx \\& \\& \therefore x^g = t \: , \: x = t^{\frac{1}{g}} \Longrightarrowdx = \dfrac{t^{\frac{1}{g}-1}}{g} dt \\& \\& \therefore \int_0^\infty \dfrac{(x)^{gn-2}}{(1+x^g)^n} dx \Longrightarrow\int_0^\infty

Integral of (1+[x^2])^(-n) from 0 to ∞

\begin{align*} & \int_0^{\infty} \dfrac{1}{(1+x^2)^n} dx \\& \\& \\& {\rm{Lemma \: (1)}}: \int_0^{\infty} \left[ f(x) \right] dx = \int_0^{\infty} \left[ \dfrac{f(\frac{1}{x})}{(x^2)} \right] dx \\& \\& {\rm{Lemma \: (2)}}: \int_0^{\frac{\pi}{2}} \left[\sin(x)\right]^{2g-1} \left[\cos(x)\right]^{2h-1}

200m MLR Estimate

16 April 2019 Formula (19/718 sampled) \begin{align*} \rm{200m} &\approx \rm{100m} \left[ 1.997383 + \left( \dfrac{7.633557}{\rm{60m}} \right) \right] - 11.1841 \pm 0.3 \\& \approx\rm{100m} \left[ 2 + \left( \dfrac{7.65}{\rm{60m}} \right) \right]

I'm an INTJ - A

Sid ... INTJ's (referred to as "Architects" and "The Master Mind") are a very rare personality type. They constitute ~ 2.1% of the populace (in regards to males ~3.3%). I

Integral of [a(x^n)+b]/[c(x)+d]

\begin{align*}I = \displaystyle \int_\alpha^\beta\dfrac{a(x)^n+b}{cx+d} \: dx &= \displaystyle \dfrac{a}{c} \int_\alpha^\beta\dfrac{(x)^n+ \frac{b}{a}}{x+\frac{d}{c}} \: dx \\&= \displaystyle \dfrac{a}{c} \int_\alpha^\beta\dfrac{(x)^n+ \Delta_1}{x+\Delta_2} \: dx \\& \therefore \: u=x+\Delta_2 \: \Longrightarrow \: u\mid_{\alpha+\Delta_2=\Delta_3}^{\beta+\Delta_2=\Delta_4} \\& \therefore

Integral of 1/(x^n+1) from 0 to ∞

\begin{aligned} I = \int_0^\infty \dfrac{1}{(x)^n+1} dx  = \dfrac{\pi}{n\sin{( \frac{\pi}{n}} )} \end{aligned} \begin{aligned} \therefore \rho=\left[ (x)^n+1 \right]^{-1} \:;\: \rho \mid_0^1 \: \Longrightarrow d\rho=\dfrac{-n(x)^{n-1}}{\left[ (x)^n+1 \right]^2} \: dx\end{aligned} \begin{aligned} \therefore (n)^{-1} \int_0^1 \left[

Creator Spotlight - Zachary Lee

Zachary Lee "Obsessed with Integrals" https://philosophicalmath.wordpress.com/ Zachary's site is very straightforward, unlike the majority of my early work. My earliest work concerned primarily finding quality sites or online PDFs to

Integral of 1/[(x^3)+1] from 0 to ∞

I=∫∞01x3+1dx=2π31.5≈1.209199576156145 Partial Fraction Decomposition Method \begin{aligned} \int_0^{\infty} \dfrac{1}{(x+1)(x^2-x+1)} \,dx \end{aligned}.\begin{aligned} \int_0^{\infty} \dfrac{1}{3(x+1)} \,dx+ \int_0^{\infty} \dfrac{x-2}{3(x^2-x+1)} \,dx \end{aligned}.\begin{aligned} \dfrac{1}{3} \int_0^{\infty} \dfrac{1}{x+1} \,dx-\dfrac{1}{3} \int_0^{\infty} \dfrac{x-2}{x^2-x+1} \,dx \end{aligned}.\begin{aligned} \left[\dfrac{1}{3} \ln{(x+1)}\right]\mid_0^{\infty}-\dfrac{1}{6}\int_0^{\infty} \dfrac{2x-1}{x^2-x+1} \,dx+\dfrac{1}{2}\int_0^{\infty} \dfrac{1}{x^2-x+1}

Fitness Plans

Below comprises 11 fitness programs/plans I offer. Once you've purchased a plan expect an intial fitness goal and evaluation via email. - C. D. Chester All Products General Nutrition /

Other Personal Outlets

So I also publish on Medium. You can see my articles as stories over there. Here's the link: C. D. Chester's Latest Medium Posts Everything I post on this website starting

VO2 Max and its Relations

References: VO2 max by Brian Mackenzie Estimation of  VO2max from the ratio between HRmax and HRrest – the Heart Rate Ratio Method by Uth et al VO2 max from a one mile