How to compute $\int_0^\infty\frac{dx}{(x^2+1)\sqrt{x^2-x+1}}$?

Question

This integral appeared in my recent integral calculus test. I tried to make a few attempts, but none of them seemed to simplify the integral.

Attempt $1$: Substituting $x\to\tan x$,

$$I=\int_0^{\pi/2}\frac{dx}{\sqrt{\tan^2x-\tan x+1}}=\sqrt2\int_0^{\pi/2}\frac{\cos x\,dx}{\sqrt{2-\sin2x}}$$

The $\tan x$ in the denominator is creating a problem, I am not able to find a way to handle it.

Attempt $2$: Substituting $x\to\frac1x$,

$$I=\int_0^\infty\frac{x\,dx}{(x^2+1)\sqrt{x^2-x+1}}\tag1$$

I tried adding this and the original integrals to get

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

But this doesn’t seem to simplify anything. Then, I tried IBP in $(1)$ with $u=\frac1{\sqrt{x^2-x+1}}$ and $dv=\frac{x\,dx}{x^2+1}$, but it gave a more complicated integral to solve:

$$I=\frac14\int_0^\infty\frac{(2x-1)\ln(x^2+1)}{(x^2-x+1)^\frac32}dx$$

Attempt $3$: Substituting $x-\frac12\to\frac{\sqrt3}2\sinh x$,

$$I=4\int_{-\frac12\ln3}^\infty\frac{dx}{3\sinh^2x+2\sqrt3\sinh x+5}=16\int_\frac1{\sqrt3}^\infty\frac{x\,dx}{3x^4+4\sqrt3x^3+14x^2-4\sqrt3x+3}$$

Attempt $4$: Euler substitution $t_\pm=\sqrt{x^2-x+1}\pm x$,

$$I=2\int_1^\infty\frac{2t_+-1}{t_+^4+2t_+^2-4t_++2}dt_+=2\int_{-1/2}^1\frac{2t_-+1}{t_-^4+2t_-^2+4t_-+2}dt_-$$

The polynomials $3x^4+4\sqrt3x^3+14x^2-4\sqrt3x+3, x^4+2x^2\pm4x+2$ have all roots unreal, so if I resort to partial fraction decomposition, it will be really messy and tedious and not something I’d be able to do in a timed test.

Any ideas as to how to tackle this integral in some other elegant way or proceed with any of the integrals I obtained in my attempts are highly appreciated.

Answer 1

Continuing from your first approach, note that $2-\sin 2x=1+(1-\sin 2x)=1+(\sin x-\cos x)^2$. Then,

$$\begin{aligned} I&=\sqrt{2}\int_0^\frac{\pi}{2}\frac{\cos x dx}{\sqrt{2-\sin 2x}}\\ &=\sqrt{2}\int_0^\frac{\pi}{2}\frac{\sin x dx}{\sqrt{2-\sin 2x}}\\ &=\frac1{\sqrt{2}}\int_0^\frac{\pi}{2}\frac{(\sin x+\cos x) dx}{\sqrt{1+(\sin x-\cos x)^2}}\\ &=\frac1{\sqrt{2}}\int_{-1}^1\frac{du}{\sqrt{1+u^2}}\\ &=\sqrt{2}\operatorname{arcsinh}(1)=\sqrt{2}\log(1+\sqrt{2}). \end{aligned}$$

Answer 2

$$\int_0^\infty\frac{dx}{(x^2+1)\sqrt{x^2-x+1}}=\int_0^\infty\frac{dx}{\sqrt x\, x\,(x+\frac1x)\sqrt{x+\frac1x-1}}$$ $$=2\int_0^\infty\frac1{\sqrt{(t-\frac1t)^2+1}\,\big((t-\frac1t)^2+2\big)}\frac{dt}{t^2}\overset{x=\frac1t}{=}2\int_0^\infty\frac{dx}{\sqrt{(x-\frac1x)^2+1}\,\big((x-\frac1x)^2+2\big)}$$ Using Glasser's master theorem

$$=2\int_0^\infty\frac{dx}{\sqrt{x^2+1}\,(x^2+2)}\overset{x=\sinh t}{=}2\int_0^\infty\frac{dt}{\sinh^2t+2}$$ $$\overset{e^{-2t}=x}{=}4\int_0^\infty\frac{dx}{x^2+6x+1}=\frac1{\sqrt2}\ln\frac{(2-\sqrt2)(3+2\sqrt2)}{(2+\sqrt2)(3-2\sqrt2)}$$ $$=\sqrt2\ln(1+\sqrt2)=1.24645...$$

Answer 3

As suggested by Quý Nhân,

$$\begin{align}\int_0^\infty\frac{dx}{(x^2+1)\sqrt{x^2-x+1}}&=2\int_0^1\frac{dy}{(y^2+1)\sqrt{3y^2+1}}\\&=2\int_0^{\pi/4}\frac{\cos z\,dz}{\sqrt{2\sin^2z+1}}\\&=\sqrt2\ln(\sqrt2+1)\end{align}$$

Answer 4

Let $\mathcal{I}$ be the integral in question. Using $x=\frac{\sqrt{3}+\sinh u}{\sqrt{3}-\sinh u}$, we get

$$ \require{cancel} \begin{align} \mathcal{I} &= \int_{0}^{\infty}\frac{1}{\left(1+x^{2}\right)\sqrt{x^{2}-x+1}}\,dx \\ &= 2\int_{-\ln\left(2+\sqrt{3}\right)}^{\ln\left(2+\sqrt{3}\right)}\frac{\sqrt{3}-\sinh u}{\cosh\left(2u\right)+5}\,du \\ &= 2\sqrt{3}\int_{-\ln\left(2+\sqrt{3}\right)}^{\ln\left(2+\sqrt{3}\right)}\frac{1}{\cosh2u+5}\,du-2\cancelto{0}{\int_{-\ln\left(2+\sqrt{3}\right)}^{\ln\left(2+\sqrt{3}\right)}\frac{\sinh u}{\cosh2u+5}\,du}\,. \\ \end{align} $$

Notice that the second integral equals $0$ because its integrand is odd. As for the other integral, use $u=\operatorname{artanh}v$ to get

$$ \int_{-\ln\left(2+\sqrt{3}\right)}^{\ln\left(2+\sqrt{3}\right)}\frac{1}{\cosh2u+5}\,du = \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}\frac{1}{6-4v^{2}}\,dv = \frac{1}{2\sqrt{6}}\left[\operatorname{artanh}\left(\sqrt{\frac{2}{3}}v\right)\right]_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{6}}\operatorname{artanh}\left(\frac{1}{\sqrt{2}}\right)\,. $$

Therefore,

$$ \bbox[0px,#95dbfe,border:3px solid #6493aa]{\bbox[10px,#ffffff,border:3.3px inset #95dbfe]{\int_{0}^{\infty}\frac{1}{\left(1+x^{2}\right)\sqrt{x^{2}-x+1}}dx = \sqrt{2}\operatorname{arcoth}\left(\sqrt{2}\right)}} $$

and we're done! o(￣︶￣)o

Answer 5

Without any change of variable, the antiderivative $$I=\int\frac{dx}{(x^2+1)\sqrt{x^2-x+1}}$$ is not difficult if you write it $$I=\int\frac{dx}{(x^2+1)\sqrt{(x-a)(x-b)}}$$ where $$(a,b)=\frac{1\pm i \sqrt{3}}{2}$$ The result is two arc tangents of complex arguments. Replacing $a$ and $b$ by their values and expanding as series for large $t$ $$J=\int_0^t \frac{dx}{(x^2+1)\sqrt{x^2-x+1}}$$ $$J=\frac{\log \left(3+2\sqrt{2}\right)}{\sqrt{2}}-\frac{1}{2 t^2}-\frac{1}{6t^3}+\frac{9}{32 t^4}+\frac{3}{16 t^5} +O\left(\frac{1}{t^6}\right)$$

The only "problem" was to recognize that $$2 \tan ^{-1}\left(\sqrt{2}\right)+(1+i) \coth ^{-1}\left(\frac{2}{\sqrt{4+3i}}\right)+$$ $$(1-i) \coth ^{-1}\left(\frac{2}{5} \sqrt{4+3 i}\right)$$ is equal to $\log \left(3+2\sqrt{2}\right)$. But this is almost immediate using the logarithmic representation of the hyperbolic arc cotangent and arc tangent functions.

$J$ is a very good approxiamtion of the exact integral : the relative error is $0.10$% if $t=2.35$ and $0.01$% if $t=3.37$.