Calculate $\int_0^\infty {\frac{x}{{\left( {x + 1} \right)\sqrt {4{x^4} + 8{x^3} + 12{x^2} + 8x + 1} }}dx}$

Question

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.

Answer 1

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.

Answer 2

Here is a method that does not require a computer.

Consider the elliptic curve $y^2=4x^4+8x^3+12x^2+8x+1$. The integrand $\frac{x\,dx}{(x+1)y}$ has poles at $x=-1$ and infinity. So we find quadratic polynomials (which already passes through the poles at infinity) with high tangency to the curve at $(-1,1)$ or at infinity. In this way, we find $$\begin{aligned}y^2&=(2x^2+8x+5)^2-24(x+1)^3\\&=4(x^2+x+1)^2+3.\end{aligned}$$. They give good substitutions: $$\begin{aligned}\frac{d}{dx}\,\mathrm{arctanh}\frac{2x^2+8x+5}{y}&=\bigg(x-1+\frac{3}{2(x+1)}\bigg)\frac1y,\\\frac{d}{dx}\,\mathrm{arctanh}\frac{2(x^2+x+1)}{y}&=\frac{4x+2}{y}.\end{aligned}$$ Combining them, we get the desired antiderivative. $$\int\frac{x\,dx}{(x+1)y}=\frac16\mathrm{arctanh}\frac{2(x^2+x+1)}{y}-\frac23\mathrm{arctanh}\frac{2x^2+8x+5}{y}+C.$$ Now just plug in values.

Answer 3

$\color{green}{\textbf{Version of 22.01.25}}$

Let $\quad x=\dfrac {1-y}{2y-1},\;$ then $$\begin{align} &I=\int\limits_0^\infty \dfrac{x\,\text dx}{(x+1)\sqrt{4x^4+8x^3+12x^2+8x+1}} =\int\limits_{\frac12}^1 \dfrac{(1-y)\,\text dy}{y\sqrt{1-12(y-y^2)^2}}\\[4pt] &=\sum\limits_{n=0}^\infty \dbinom{-\frac12}{n}\, 12^n \int\limits_{\frac12}^1 (y-1)^{2n+1}y^{2n-1}\,\text dy, \end{align}$$ $$I=J_0(1)-J_0\left(\dfrac12\right)+\sum\limits_{n=1}^\infty \left(\dfrac32\right)_{(n-1)} \dfrac{12^n}{2n!}\,\left(J_n(1)-J_n\left(\dfrac12\right)\right),\tag1$$ where $$J_0(y)=\int\dfrac{1-y}{y}\,\text dy =\ln y -y,$$ $$I_0=J_0(1)-J_0\left(\dfrac12\right)=\ln2-\dfrac12\approx 0.19314718,\tag2$$ At the same time, basing on the properties of Beta function, Incomplete Beta function and Regularized Beta function, we can get

$$\begin{align} &J_n(1)-J_n\left(\dfrac12\right)\bigg|_{n>0} =\int\limits_0^1 y^{2n-1}(y-1)^{2n+1}\,\text dy -\int\limits_0^{\frac12} y^{2n-1}(y-1)^{2n+1}\text dy\\[4pt] &= \operatorname{B}\big(2n, 2n+2 \big) - \operatorname{B}_{\frac12}\big(2n, 2n+2 \big) = \operatorname{B}\big(2n, 2n+2 \big) \left(1-\operatorname{I}_{\frac12}\big(2n, 2n+2 \big)\right)\\[4pt] &=\operatorname{B}_{\frac12}\big(2n+2, 2n \big) =\operatorname{B}\big(2n+2, 2n \big)\sum\limits_{j=2n+2}^{4n+1} \dbinom{4n+1}{j}\left(\dfrac12\right)^j\left(1-\dfrac12\right)^{4n+1-j}\\[4pt] &=\operatorname{B}\big(2n+2, 2n \big) \dfrac1{2^{4n+1}} \sum\limits_{j=2n+2}^{4n+1}\dbinom{4n+1}{j} =\operatorname{B}\big(2n+2, 2n\big)\left(\dfrac12-\dfrac1{2^{4n+1}}\dbinom{4n+1}{2n+1}\right)\\[4pt] &=\dfrac12\,\operatorname{B}\big(2n+2, 2n\big) -\dfrac{\Gamma(2n+2)\,\Gamma(2n)}{\Gamma(4n+2)}\cdot \dfrac1{2^{4n+1}}\dfrac{\Gamma(4n+2)}{\Gamma(2n+2)\Gamma(2n+1)}\\[4pt] &=\dfrac12\,\operatorname{B}\big(2n+2, 2n\big)- \dfrac1{\,2^{4n+2}n}. \end{align}$$

Then $$\begin{align} &I_1=-\sum\limits_{n=1}^\infty \left(\dfrac32\right)_{(n-1)} \dfrac{12^n}{2n!}\,\left(\dfrac1{2^{4n+2}\,n}\right) =-\sum\limits_{n=1}^\infty \dfrac1{8n!\,n}\,\left(\dfrac32\right)_{(n-1)} \left(\dfrac34\right)^n,\\[4pt] &I_1=-\dfrac12 \ln\,\left(\dfrac43\right)\approx -0.143841836;\tag3\\[4pt] \end{align}$$ (see also WA result) $$\begin{align} &I_2=\sum\limits_{n=1}^\infty \left(\dfrac32\right)_{(n-1)} \dfrac{12^n}{4n!}\,\operatorname{B}\big(2n+2, 2n\big)=\dfrac3{20}\operatorname{_4F_3}\left(1, 1, \dfrac32, \dfrac52; \dfrac74, 2, \dfrac94; \dfrac34\right),\\[4pt] &I_2\approx 0.26895094, \end{align}$$ (see also WA calculations),

wherein $\dfrac32-\dfrac9{20}\operatorname{_4F_3}\left(1, 1, \dfrac32, \dfrac52; \dfrac74, 2, \dfrac94; \dfrac34\right)$ and $\;\ln2\;$ have more than $\;180\;$ identical decimal signs.

Therefore, we have the result in a closed form, which corresponds with the value $$I=I_0+I_1+I_2=\ln2-\dfrac12-\dfrac12\ln\dfrac43+\dfrac12-\dfrac13 \ln2 = \dfrac12 \ln3 -\dfrac13 \ln2. $$

OLD VERSION

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

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

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

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

Answer 4

This partial answer piggybacks on Sangchul Lee's earlier suggested transformations to show that, with a few more steps, we can indeed reveal some elliptic integrals. (Applying the fundamental theorem of calculus is all that's left to do, but I'm not totally familiar with EIs, and WolframAlpha is giving back a potentially bogus result when plugging in the lower limit of the $u$-integral below.)

Let $a=\dfrac2{\sqrt3}$, $b=\sqrt{1+a}$, and $k=\sqrt{\dfrac{2a}{1+a}}$.

$$\begin{align*} \mathcal I &= \int_0^\infty \frac x{x+1} \cdot \frac{dx}{\sqrt{4x^4+8x^3+12x^2+8x+1}} \\ &= \int_0^1 \left(\frac1{2\sqrt{1-y}} - \frac1{1+\sqrt{1-y}}\right) \frac{dy}{\sqrt{4-3y^2}} & y=\frac{4x(1+x)}{2x^2+2x+1} \\ &= \frac a4 \int_0^\tfrac\pi3 \frac{1-\sqrt{1-a\sin t}}{1+\sqrt{1-a\sin t}} \cdot \frac{dt}{\sqrt{1-a \sin t}} & y=a\sin t \\ &= \frac a2 \int_\tfrac\pi4^\tfrac{5\pi}{12} \frac{1-\sqrt{1+a\cos(2u)}}{1+\sqrt{1+a\cos(2u)}} \cdot \frac{du}{\sqrt{1+a\cos(2u)}} & u=\frac t2+\frac\pi4 \\ &= \frac12 \int_\tfrac\pi4^\tfrac{5\pi}{12} \frac{2-b\sqrt{1-k^2\sin^2u} - \frac1{b\sqrt{1-k^2\sin^2u}}}{1-2\sin^2u} \, du \\ &= \frac1k I_1 - \frac b{2k} I_2 - \frac1{2bk} I_0 & \sin u=\frac1k \sin v \end{align*}$$

where

$$I_n = \int_{\arcsin\left(\sqrt3-1\right)}^\tfrac\pi2 \frac{(\cos v)^n}{\left(1-\frac2{k^2}\sin^2v\right) \sqrt{1-\frac1{k^2}\sin^2v}} \, dv.$$

$I_0$ matches the IEIot3K $(\Pi)$, and $I_2$ may be evaluated recursively and in terms of the IEIot1K $(F)$:

$$\begin{align*} I_2 &= \int\frac{1-\sin^2v}{\left(1-\frac2{k^2}\sin^2v\right) \sqrt{1-\frac1{k^2}\sin^2v}} \, dv \\ &= \left(1-\frac{k^2}2\right) I_0 + \frac{k^2}2 \int\frac{dv}{\sqrt{1-\frac1{k^2}\sin^2v}} \end{align*}$$

$I_1$ has an elementary antiderivative by way of substituting $\sinh w=\sqrt{1-\dfrac1{k^2}}\tan v$:

$$\begin{align*} & \int \frac{d(\sin v)}{\left(1-\frac2{k^2}\sin^2v\right) \sqrt{1-\frac1{k^2}\sin^2v}} \\ &= \int \frac{d(\tan v)}{\left(1+\left(1-\frac2{k^2}\right)\tan^2v\right) \sqrt{1+\left(1-\frac1{k^2}\right)\tan^2v}} \\ &= \int \frac{\sqrt{1-\frac1{k^2}}}{1-\frac1{k^2} + \left(1-\frac2{k^2}\right) \sinh^2w} \, dw \\ &= \int \frac{\sqrt{1-\frac1{k^2}} \, d(\tanh w)}{1-\frac1{k^2}-\frac1{k^2}\tanh^2w} \end{align*}$$