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

Question

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

My Attempt:

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

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

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

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},\quad dx=-\frac{2}{\left( t+1\right) ^{2}}dt,\quad t=-% \frac{x-1}{x+1} \end{equation*} reduces the given integral \begin{equation*} I=\int \frac{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}}dt-2\int \frac{1}{ \left( 3t^{2}+1\right) \sqrt{t^{2}+3}}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}}dt=-2\int \frac{1}{-8+3u^{2}}\,du,\qquad u=\sqrt{t^{2}+3} \\ &=&\frac{\sqrt{6}}{6}\operatorname{arctanh}(\frac{\sqrt{6}}{4}u) \\ I_{2} &=&-2\int \frac{1}{\left( 3t^{2}+1\right) \sqrt{t^{2}+3}}dt=-2\int \frac{1}{8v^{2}+1}\,dv,\qquad v=\frac{t}{\sqrt{t^{2}+3}} \\ &=&-\frac{\sqrt{2}}{2}\arctan (2\sqrt{2}v) \\ I &=&I_{1}+I_{2}=\frac{\sqrt{6}}{6}\operatorname{arctanh}\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{arctanh}\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.