Compute the integral $\int\frac{\mathrm dx}{(x^2-x+1)\sqrt{x^2+x+1}}$

Question

Compute the indefinite integral $$ \int\frac{\mathrm dx}{(x^2-x+1)\sqrt{x^2+x+1}} $$

My Attempt:

$$ \int\frac{\mathrm dx}{(x^2-x+1)\sqrt{x^2+x+1}}=\int\frac{1}{(x^2-x+1)^{3/2}.\sqrt{\dfrac{x^2+x+1}{x^2-x+1}}}\mathrm dx$$

Now define $t$ such that $t^2=\dfrac{x^2+x+1}{x^2-x+1}$ to get

$$ \begin{align} 2t\mathrm dt& = \frac{(x^2-x+1)(2x+1)-(x^2+x+1)\cdot (2x-1)}{(x^2-x+1)^2}\mathrm dx\\ \implies2t\mathrm dt &= \frac{-4x^2+2x+2}{(x^2-x+1)^2}\mathrm dx \end{align} $$

I don't know how to proceed from here.

Answer 1

Instead of the Euler substitution, we can use the following technique adapted from a general procedure explained in 2.252 of Table of Integrals, Series, and Products by I.S. Gradshteyn and I.M. Ryzhik, 7th. ed.

The substitution \begin{equation*} x=\frac{1-t}{t+1},\mathrm dx=-\frac{2}{\left( t+1\right) ^{2}}\mathrm dt,\quad t=-% \frac{x-1}{x+1} \end{equation*} reduces the given integral \begin{equation*} I=\int \frac{\mathrm dx}{(x^{2}-x+1)\sqrt{x^{2}+x+1}} \end{equation*} to the sum of the two integrals \begin{equation*} I=-2\int \frac{t}{\left( 3t^{2}+1\right) \sqrt{t^{2}+3}}\mathrm dt-2\int \frac{1}{ \left( 3t^{2}+1\right) \sqrt{t^{2}+3}}\mathrm dt. \end{equation*}

The first integral can be evaluated by the substitution \begin{equation*} u=\sqrt{t^{2}+3}, \end{equation*} while the second one is integrable by the substitution \begin{equation*} v=\frac{t}{\sqrt{t^{2}+3}}. \end{equation*} Both substitutions transform the integrands into simple rational fractions as follows \begin{eqnarray*} I_{1} &=&-2\int \frac{t}{\left( 3t^{2}+1\right) \sqrt{t^{2}+3}}\mathrm dt=-2\int \frac{1}{-8+3u^{2}}\mathrm du,\qquad u=\sqrt{t^{2}+3} \\ &=&\frac{\sqrt{6}}{6}\operatorname{arccoth}\left(\frac{\sqrt{6}}{4}u\right)+C\\ I_{2} &=&-2\int \frac{1}{\left( 3t^{2}+1\right) \sqrt{t^{2}+3}}\mathrm dt=-2\int \frac{1}{8v^{2}+1}\mathrm dv,\qquad v=\frac{t}{\sqrt{t^{2}+3}} \\ &=&-\frac{\sqrt{2}}{2}\arctan\left(2\sqrt{2}v\right)+C\\ I &=&I_{1}+I_{2}=\frac{\sqrt{6}}{6}\operatorname{arccoth}\frac{\sqrt{6}\sqrt{t^{2}+3} }{4}-\frac{\sqrt{2}}{2}\arctan \frac{2\sqrt{2}t}{\sqrt{t^{2}+3}}+C. \end{eqnarray*} We finally get

\begin{equation*} I=\frac{\sqrt{6}}{6}\operatorname{arccoth}\frac{\sqrt{6}\sqrt{x^{2}+x+1}}{2\left( x+1\right) }+\frac{\sqrt{2}}{2}\arctan \frac{\sqrt{2}\left( x-1\right) }{ \sqrt{x^{2}+x+1}}+C. \end{equation*}

Answer 2

According to http://en.wikipedia.org/wiki/Euler_substitution, this integral can have these four approaches to solve:

Approach $1$:

Let $u=x+\sqrt{x^2+x+1}$ ,

Then $x=\dfrac{u^2-1}{2u+1}$

$dx=\dfrac{2u(2u+1)-(u^2-1)2}{(2u+1)^2}du=\dfrac{2u^2+2u+2}{(2u+1)^2}du$

$\therefore\int\dfrac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}$

$=\int\dfrac{\dfrac{2u^2+2u+2}{(2u+1)^2}}{\left(\left(\dfrac{u^2-1}{2u+1}\right)^2-\dfrac{u^2-1}{2u+1}+1\right)\left(u-\dfrac{u^2-1}{2u+1}\right)}du$

$=\int\dfrac{\dfrac{2u^2+2u+2}{(2u+1)^2}}{\dfrac{(u^2-1)^2-(u^2-1)(2u+1)+(2u+1)^2}{(2u+1)^2}\times\dfrac{u^2+u+1}{2u+1}}du$

$=2\int\dfrac{2u+1}{u^4-2u^3+u^2+6u+3}du$

Other approaches are similar.

Answer 3

The integral can be rewritten as

$$\mathcal I=\int\frac{\mathrm dx}{(x^2-x+1)\sqrt{x^2+x+1}}=\int\frac{\mathrm dx}{|x|^\frac32\left(|x|+\frac1{|x|}-\text{sgn}(x)\right)\sqrt{|x|+\frac1{|x|}+\text{sgn}(x)}}$$

This can be simplified by the substitution $t=\frac1{\sqrt{|x|}}$.

$$\mathcal I=-\text{sgn}(x)\int\frac{2\mathrm dt}{\left(t^2+\frac1{t^2}-\text{sgn}(x)\right)\sqrt{t^2+\frac1{t^2}+ \text{sgn}(x)}}$$

Now, writing the numerator of the integrand as $\left(1+\frac1{t^2}\right)+\left(1-\frac1{t^2}\right)$, we get:

$$\mathcal I=-\text{sgn}(x)\underbrace{\int\frac{\mathrm d\left(t-\frac1t\right)}{\left(\left(t-\frac1t\right)^2+2-\text{sgn}(x)\right)\sqrt{\left(t-\frac1t\right)^2+2+\text{sgn}(x)}}}_{\mathcal I_1}-\text{sgn}(x)\underbrace{\int\frac{\mathrm d\left(t+\frac1t\right)}{\left(\left(t+\frac1t\right)^2-2- \text{sgn}(x)\right)\sqrt{\left(t+\frac1t\right)^2-2+\text{sgn}(x)}}}_{\mathcal I_2}$$

It is imminent that we need to make $2$ cases to handle $\text{sgn}(x)$:$-$

$\textbf{Case 1, $x>0\,$:}$

$$\mathcal I=-\underbrace{\int\frac{\mathrm d\left(t-\frac1t\right)}{\left(\left(t-\frac1t\right)^2+1\right)\sqrt{\left(t-\frac1t\right)^2+3}}}_{\mathcal I_1}-\underbrace{\int\frac{\mathrm d\left(t+\frac1t\right)}{\left(\left(t+\frac1t\right)^2-3\right)\sqrt{\left(t+\frac1t\right)^2-1}}}_{\mathcal I_2}$$

These integrals can be evaluated in any of the following ways:

$1.$ For $\mathcal I_1$, substitute

• $t-\frac1t=\frac1u$
• $t-\frac1t=\tan\theta$
• $t-\frac1t=\sqrt3\sinh u$

$2.$ For $\mathcal I_2$, substitute

• $t+\frac1t=\frac1u$
• $t+\frac1t=\sqrt3\coth\theta$
• $t+\frac1t=\cosh u$

$\textbf{Case 2, $x<0\,$:}$

$$\mathcal I=\underbrace{\int\frac{\mathrm d\left(t-\frac1t\right)}{\left(\left(t-\frac1t\right)^2+3\right)\sqrt{\left(t-\frac1t\right)^2+1}}}_{\mathcal I_1}+\underbrace{\int\frac{\mathrm d\left(t+\frac1t\right)}{\left(\left(t+\frac1t\right)^2-1\right)\sqrt{\left(t+\frac1t\right)^2-3}}}_{\mathcal I_2}$$

These integrals can be evaluated in any of the following ways

$1.$ For $\mathcal I_1$, substitute

• $t-\frac1t=\frac1u$
• $t-\frac1t=\sqrt3\tan\theta$
• $t-\frac1t=\sinh u$

$2.$ For $\mathcal I_2$, substitute

• $t+\frac1t=\frac1u$
• $t+\frac1t=\coth\theta$
• $t+\frac1t=\sqrt3\cosh u$

The final result comes out to be

$$\boxed{\frac1{\sqrt2}\sin^{-1}\left(\sqrt{\frac23}\frac{x-1}{\sqrt{x^2-x+1}}\right)+\frac1{\sqrt6}\tanh^{-1}\left(\sqrt{\frac23}\frac{x+1}{\sqrt{x^2+x+1}}\right)+C}$$