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How does the PBR-Blend material (beta) work in Blender 2.72?

The new material mentioned in the question “How does the PBR-Blend material (beta) work in Blender 2.72?” is very interesting, but it’s still beta.
Does anyone know how it is supposed to work?

I guess I understand what it does, but I want to know more detailed information about how it works.

A:

TL;DR
It seems to use DXT compression internally.
The Details
According to the Blender 2.7x Material Documentation, the PBR-Blend Material is a new internal material type (not just a new material) and not just a remapping of other material types:

The PBR-Blend material is a new internal material type.
It has an internal variable called “PBR_Blend” that is a
blend of diffuse and specular terms.

Not much else seems to be documented about it, as of today. Theoretically it might use some post-processing for out-of-range settings.
Due to it being internal it does not use the standard material system but is uses something different and less well documented.
One thing to note is that the PBR-Blend material is Blender’s version of Unreal’s PBR-GLSL material found here:

U-SHADOW MATERIALS
To mimic the shader hardware we could use
PBR, GLSL, or both

https://colab.research.google.com/drive/19xDSSimsQs4aa5g_MmiBep8aTuQt6xeP
https://colab.research.google.com/drive/1EME4MVQEoholTbqr1blRvooF1mJBveWF
https://colab.research.google.com/drive/1mUOMqV3p_PtxxtCR6y6gFwCVvuOL-QFO
https://colab.research.google.com/drive/1hP58ny_qWcD1JKEQyG9tgVdwaWGAEqui
https://ello.co/duccavaters/post/k3czwjhm11qfw-diolizaq
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https://colab.research.google.com/drive/1sbWdJCQVrox9mVB5nA_YVPvnlEhzjJ6X
https://ello.co/1liagregxcomppu/post/g3hwvvwj3cbdjz3qjrd3lw
https://documenter.getpostman.com/view/21830446/UzXUQaDG
https://colab.research.google.com/drive/1mCyi-blsh5oJySwm-wpXUXmw93KBy1eZ

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MD5 Hash : 158fb5ebd8719b4f5330b75325e456db

Description : Systemcare Ultimate 8 now supports Windows 10!
Systemcare Ultimate 8 now supports Windows 10!Q:

Homework questions on Differential Geometry

I am currently doing some computer experiments on Differential Geometry and I have some questions :
1)
$$ \iint_R\Big[\frac {x^3-6x+4}{x^2-x+1} -\frac {x-2}{(x-1)^2}\Big]dxdy$$
I have no idea how to approach the problem, is there any formula to solve it?
2) The above integral has two solutions x = 1 or x = 0 for equation $y^2 = x(x-1)$
why is this?
Thanks!

A:

$
ewcommand{\Reals}{\mathbf{R}}
ewcommand{\Cpx}{\mathbf{C}}$The given integral is $$\begin{align*} \iint_R\left(\frac{x^3-6x+4}{x^2-x+1} – \frac{x-2}{(x-1)^2}\right)\,dx\,dy&=\iint_R \frac{x^3-6x+4}{x^2-x+1}\,dx\,dy – \iint_R\frac{x-2}{(x-1)^2}\,dx\,dy\tag{1}\\ &=\int_0^1 \frac{1}{x}dx-\int_1^2 \frac{2}{x-1}\,dx\tag{2}\\ &=\frac{\ln 2}{2}.\tag{
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