14

Prove $$I=\int_0^\infty {\frac{x}{{\left( {x + 1} \right)\sqrt {4{x^4} + 8{x^3} + 12{x^2} + 8x + 1} }}dx} = \frac{{\ln 3}}{2} - \frac{{\ln 2}}{3}.$$

First note that $$4{x^4} + 8{x^3} + 12{x^2} + 8x + 1 = 4{\left( {{x^2} + x + 1} \right)^2} - 3,$$ we let $${x^2} + x + 1 = \frac{{\sqrt 3 }}{{2\cos \theta }} \Rightarrow x = \sqrt { - \frac{3}{4} + \frac{{\sqrt 3 }}{{2\cos \theta }}} - \frac{1}{2},$$ then $$I=\frac{1}{2}\int_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{\left( {\sqrt {2\sqrt 3 \sec \theta - 3} - 1} \right)\sec \theta }}{{\left( {\sqrt {2\sqrt 3 \sec \theta - 3} + 1} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }}d\theta } .$$ we have \begin{align*} &\frac{{\left( {\sqrt {2\sqrt 3 \sec \theta - 3} - 1} \right)\sec \theta }}{{\left( {\sqrt {2\sqrt 3 \sec \theta - 3} + 1} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }} = \frac{{{{\left( {\sqrt {2\sqrt 3 \sec \theta - 3} - 1} \right)}^2}\sec \theta }}{{\left( {2\sqrt 3 \sec \theta - 4} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }}\\ =& \frac{{\left( {2\sqrt 3 \sec \theta - 2 - 2\sqrt {2\sqrt 3 \sec \theta - 3} } \right)\sec \theta }}{{\left( {2\sqrt 3 \sec \theta - 4} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }} = \frac{{\left( {\sqrt 3 \sec \theta - 1 - \sqrt {2\sqrt 3 \sec \theta - 3} } \right)\sec \theta }}{{\left( {\sqrt 3 \sec \theta - 2} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }}\\ = &\frac{{\left( {\sqrt 3 \sec \theta - 1} \right)\sec \theta }}{{\left( {\sqrt 3 \sec \theta - 2} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }} - \frac{{\sec \theta }}{{\sqrt 3 \sec \theta - 2}}. \end{align*} and $$\int {\frac{{\sec \theta }}{{\sqrt 3 \sec \theta - 2}}d\theta } = \ln \frac{{\left( {2 + \sqrt 3 } \right)\tan \frac{\theta }{2} - 1}}{{\left( {2 + \sqrt 3 } \right)\tan \frac{\theta }{2} + 1}}+ C.$$ while \begin{align*}&\int {\frac{{\left( {\sqrt 3 \sec \theta - 1} \right)\sec \theta }}{{\left( {\sqrt 3 \sec \theta - 2} \right)\sqrt {2\sqrt 3 \sec \theta - 3} }}d\theta } = \int {\frac{{\sqrt 3 - \cos \theta }}{{\left( {\sqrt 3 - 2\cos \theta } \right)\sqrt {2\sqrt 3 \cos \theta - 3{{\left( {\cos \theta } \right)}^2}} }}d\theta } \\ = &\frac{1}{2}\int {\frac{1}{{\sqrt {2\sqrt 3 \cos \theta - 3{{\left( {\cos \theta } \right)}^2}} }}d\theta } + \frac{{\sqrt 3 }}{2}\int {\frac{1}{{\left( {\sqrt 3 - 2\cos \theta } \right)\sqrt {2\sqrt 3 \cos \theta - 3{{\left( {\cos \theta } \right)}^2}} }}d\theta } . \end{align*} But how can we continue? It is related to elliptic integral.

Eufisky
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  • In case others reading this question were planning to check: Mathematica can't verify the result answer symbolically, but it appears to check out numerically. – Semiclassical Feb 03 '17 at 14:35
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    My bet is on a straightforward consequence of Glasser's master theorem (http://mathworld.wolfram.com/GlassersMasterTheorem.html). But I am too lazy to perform the involved computations. I slayed a similar beast (http://math.stackexchange.com/questions/266181/a-nasty-integral-of-a-rational-function) once without invoking GMT, and that was painful enough. – Jack D'Aurizio Feb 03 '17 at 14:47
  • @JackD'Aurizio i second your remarks – tired Feb 03 '17 at 17:05
  • Here are some other representations that might be useful for further computation: \begin{align} I &= \int_{0}^{1} \frac{u}{\sqrt{u^4 + 4u^3 - 6u^2 + 4u + 1}} , du \ &= \int_{0}^{1} \left( \frac{1}{2\sqrt{1-y}} - \frac{1}{1+\sqrt{1-y}} \right) \frac{dy}{\sqrt{4 - 3y^2}}. \end{align} – Sangchul Lee Feb 04 '17 at 02:54

2 Answers2

5

This is a pseudo-elliptic integral, it has an elementary anti-derivative:

$$\int \frac{x}{(x+1)\sqrt{4x^4+8x^3+12x^2+8x+1}} dx = \frac{\ln\left[P(x)+Q(x)\sqrt{4x^4+8x^3+12x^2+8x+1}\right]}{6} - \ln(x+1) + C$$

where $$P(x) = 112x^6+360x^5+624x^4+772x^3+612x^2+258x+43$$ and $$Q(x) = 52x^4+92x^3+30x^2-22x-11$$

To obtain this answer, just follow the systematic method of symbolic integration over simple algebraic extension. Alternatively, you can throw it to a CAS with Risch algorithm implemented (not Mathematica), a convenient software is the online Axiom sandbox.

pisco
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0

HINT 1

$$x=2x+1-(x+1),$$ so $$I=\frac12\int_0^\infty\frac{(2x+1)dx}{(x+1)\sqrt{(x^2+x+1)^2-\frac34}} - \frac12\int_0^\infty\frac{(2x+1)dx}{(2x+1)\sqrt{(x^2+x+1)^2-\frac34}}.$$

HINT 2

$$(2x+1)dx = d(x^2+x+1) = \frac{\sqrt3}2d\sec\theta$$

Yuri Negometyanov
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