Indefinite Integral (06)

[jacks]

Find \[\int\frac{1}{\left(x^2-x+1\right)\sqrt{x^2+x+1}}dx\]

[Grigorios Kostakos]

We make the substitution \(t=\dfrac{x-1}{x+1}\quad\Rightarrow\quad x=\dfrac{1+t}{1-t}\,,\; dx=\dfrac{2}{(1-t)^2}\,dt\,.\)\begin{align*}
I&=\displaystyle\int {\frac{1}{(x^2-x+1)\,\sqrt{x^2+x+1}}\,dx}\\
& = \int {\frac{(1-t)^2}{(3t^2+1)\,\sqrt{\frac{t^2+3}{(1-t)^2}}}\,\dfrac{2}{(1-t)^2}\,dt}\\
& =\int {\frac{2\,(1-t)}{(3t^2+1)\,\sqrt{t^2+3}}\,dt}\\
& \mathop{=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{u\,=\,\frac{t}{\sqrt{3}}} \\
{\sqrt{3}\,du\,=\,dt} \end{subarray}}\,-\int {\frac{2\sqrt{3}\,\bigl({u\sqrt{3}-1}\bigr)}{(9u^2+1)\,\sqrt{3u^2+3}}\,du}\\
& =-\int {\frac{2\,\bigl({u\sqrt{3}-1}\bigr)}{(9u^2+1)\,\sqrt{u^2+1}}\,du}\\
& =-\int {\frac{2u\sqrt{3}}{(9u^2+1)\,\sqrt{u^2+1}}\,du}+\int {\frac{2}{(9u^2+1)\,\sqrt{u^2+1}}\,du}\\
&\mathop{=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{u\,=\,\tan\theta} \\
{du\,=\,\frac{1}{\cos^2\theta}\,d\theta}\end{subarray}}\,-\int {\frac{2\sqrt{3}\,\tan\theta}{(9\tan^2\theta+1)\,\sqrt{\tan^2\theta+1}}\,\frac{1}{\cos^2\theta}\,d\theta}+\int {\frac{2}{(9\tan^2\theta+1)\,\sqrt{\tan^2\theta+1}}\,\frac{1}{\cos^2\theta}\,d\theta}\\
&=-\int {\frac{2\sqrt{3}\,\frac{\sin\theta}{\cos\theta}}{\frac{9\sin^2\theta+\cos^2\theta}{\cos^2\theta}\,\frac{1}{\cos\theta}}\,\frac{1}{\cos^2\theta}\,d\theta}+\int {\frac{2}{\frac{9\sin^2\theta+\cos^2\theta}{\cos^2\theta}\,\frac{1}{\cos\theta}}\,\frac{1}{\cos^2\theta}\,d\theta}\\
&=-\int {\frac{2\sqrt{3}\,\sin\theta}{9\sin^2\theta+\cos^2\theta}\,d\theta}+\int {\frac{2\cos\theta}{9\sin^2\theta+\cos^2\theta}\,d\theta}\\
& =-\int {\frac{2\sqrt{3}\,\sin\theta}{9-8\cos^2\theta}\,d\theta}+\int {\frac{2\cos\theta}{8\sin^2\theta+1}\,d\theta}\\
&\mathop{=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{r\,=\,\cos\theta} \\
{s\,=\,\sin\theta}\end{subarray}}\,\int {\frac{2\sqrt{3}}{9-8r^2}\,dr}+\int {\frac{2}{8s^2+1}\,ds}\\
&=\frac{\sqrt{2}\sqrt{3}}{6}\int{\frac{1}{1-\bigl({\frac{2r\sqrt{2}}{3}}\bigr)^2}\,d\bigl({\tfrac{2r\sqrt{2}}{3}}\bigr)}+\frac{1}{\sqrt{2}}\int {\frac{2}{\bigl({2s\sqrt{2}\,}\bigr)^2+1}\,d\bigl({2s\sqrt{2}\,}\bigr)}\\
&=\frac{\sqrt{2}\sqrt{3}}{6}{\rm{arctanh}}\bigl({\tfrac{2r\sqrt{2}}{3}}\bigr)+\frac{1}{\sqrt{2}}\arctan\bigl({2s\sqrt{2}\,}\bigr)+c\\
&\mathop{=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{r\,=\,\cos\theta} \\
{s\,=\,\sin\theta} \end{subarray}}\,\frac{\sqrt{2}\sqrt{3}}{6}{\rm{arctanh}}\bigl({\tfrac{2\sqrt{2}}{3}\cos\theta}\bigr)+\frac{1}{\sqrt{2}}\arctan\bigl({2\sqrt{2}\,\sin\theta}\bigr)+c\\
&\stackrel{{u\,=\,\tan\theta}}{=\!=\!=\!=\!=}\frac{\sqrt{2}\sqrt{3}}{6}{\rm{arctanh}}\bigl({\tfrac{2\sqrt{2}}{3\sqrt{1+u^2}}}\bigr)+\frac{1}{\sqrt{2}}\arctan\bigl({\tfrac{2u\sqrt{2}}{\sqrt{1+u^2}}}\bigr)+c\\
&\stackrel{u\,=\,\frac{t}{\sqrt{3}}}{=\!=\!=\!=}\frac{\sqrt{2}\sqrt{3}}{6}{\rm{arctanh}}\bigl({\tfrac{2\sqrt{2}\sqrt{3}}{3\sqrt{3+t^2}}}\bigr)+\frac{1}{\sqrt{2}}\arctan\bigl({\tfrac{2t\sqrt{2}}{\sqrt{3+t^2}}}\bigr)+c\\
&\stackrel{{t\,=\,\frac{x-1}{x+1}}}{=\!=\!=\!=\!=}\frac{\sqrt{2}\sqrt{3}}{6}\,{\rm{arctanh}}\bigl({\tfrac{\sqrt{2}\sqrt{3}\,(x+1)}{3\sqrt{x^2+x+1}}}\bigr)+\frac{\sqrt{2}}{2}\,\arctan\bigl({\tfrac{\sqrt{2}\,(x-1)}{\sqrt{x^2+x+1}}}\bigr)+c\,.
\end{align*}