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Evaluate $$I=\int \frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}$$

My Try:

we have $x^2-x+1=(x+w)(x+w^2)$ where $w$ is complex cube root of unity

I have splitted $I$ as

$$I=AI_1+BI_2$$ where $A,B$ are some constants

$$I_1=\int \frac{dx}{(x+w)\sqrt{x^2+x+1}}$$

By taylor's series $$x^2+x+1=P(x+w)^2+Q(x+w)+R=(x+w)^2 \left(P+\frac{Q}{x+w}+\frac{R}{(x+w)^2}\right)$$ for some constants complex $P,Q,R$

hence

$$I_1=\int \frac{\frac{dx}{(x+w)^2}}{\sqrt{ \left(P+\frac{Q}{x+w}+\frac{R}{(x+w)^2}\right)}}=\int \frac{dt}{\sqrt{Rt^2+Qt+P}}$$ which is a standard Integral. Similar analysis for $I_2$.

Any other approach?

Quanto
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Ekaveera Gouribhatla
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  • $$\mathscr{I}:=\int\frac{1}{\left(x^2-x+1\right)\cdot\sqrt{x^2+x+1}}\space\text{d}x=2\int\frac{1}{\left(x^2-x+1\right)\cdot\sqrt{\left(1+2x\right)^2+3}}\space\text{d}x\tag1$$

    Now, substitute $\text{u}:=1+2x$:

    $$\mathscr{I}=4\int\frac{1}{\left(\text{u}^2+4\text{u}+7\right)\cdot\sqrt{\text{u}^2+3}}\space\text{d}\text{u}\tag2$$

    Substitute $\text{u}=\sqrt{3}\tan\left(\text{p}\right)$:

    $$\mathscr{I}=4\int\frac{\cos\left(\text{p}\right)}{5+4\sin\left(\frac{\pi}{6}-2\text{p}\right)}\space\text{d}\text{p}\tag3$$

     – Jan Eerland Sep 08 '17 at 11:46
  • I don't know if the integral will be harder but,you can try Euler substitution because your integral has the form of $\displaystyle \int R(x,\sqrt{ax^2+bx+c})dx$, where $R$ is a rational function of $x$ and $\sqrt{ax^2+bx+c}$. In your case $a=1>0$ so the substitution is $\sqrt{x^2+x+1}=x+t \implies x=\frac{1-t^2}{2t-1}$. From here you can do it. – Adrian Sep 08 '17 at 12:53

2 Answers2

3

Substituting

$$ x = \frac{\alpha + \beta t}{1 +t}$$

we have:

$$ x^2- x+1 = \frac{(\alpha +\beta t)^2-(1+t)(\alpha +\beta t)+(1+t)^2}{(1+t)^2}$$

$$ x^2+x +1 = \frac{(\alpha +\beta t)^2+(1+t)(\alpha +\beta t)+(1+t)^2}{(1+t)^2}$$

Numbers $ \alpha, \beta $ we define like, that coefficients at $ t $ are zero.

Hence

$$ 2\alpha \beta - \alpha - \beta + 2 =0,\ \ 2\alpha \beta +\alpha +\beta +2=0.$$

$$ \alpha = 1, \ \ \beta = -1.$$

We have

$$ x = \frac{1-t}{1+t}; \ \ dx = \frac{-2dt}{(1+t)^2};$$

$$ x^2 -x +1 = \frac{3t^2 +2}{1 + t^2};$$

$$\sqrt{x^2+x +1} =\frac{\sqrt{t^2+3}}{1+t}; \ \ 1+t>0. $$

$$I =-2\int \frac{(t+1)dt}{(3t^2+1)\sqrt{t^2+3}}= -2\int \frac{tdt}{(3t^2+1)\sqrt{t^2+3}} - 2\int \frac{dt}{(3t^2+1)\sqrt{t^2+3}} = I_{1} + I_{2}.$$

$$ I_{1} = -2\int \frac{tdt}{(3t^2+1)\sqrt{t^2+3}} = (\sqrt{t^2+3}= u) = 2\int \frac{du}{8 -3u^2} = \frac{1}{2\sqrt{6}}\ln \left| \frac{2\sqrt{2}+\sqrt{3}u}{2\sqrt{2}-\sqrt{3}u}\right| = \frac{1}{\sqrt{6}}\ln\left|\frac{2\sqrt{2}+\sqrt{3(t^2+3)}}{2\sqrt{2}- \sqrt{3(t^2+3)}}\right|=\frac{1}{\sqrt{6}}\ln\left|\frac{(1+x)\sqrt{2}+\sqrt{3(x^2+x+1)}}{\sqrt{x^2-x+1}}\right|.$$

$$ I_{2}=-2\int\frac{dt}{(3t^2+1)\sqrt{t^2+3}}=(\frac{t}{\sqrt{t^2+3}}=z)=-2 \int\frac{dz}{8z^2 +1} =-\frac{1}{\sqrt{2}}\arctan\left(\frac{2\sqrt{z}}{1}\right)= -\frac{1}{\sqrt{2}}\arctan\left(\frac{\sqrt{2}(1-x)}{\sqrt{x^2+x +1}}\right).$$

In the end

$$ I =\frac{1}{\sqrt{6}}\ln\left|\frac{(1+x)\sqrt{2}+\sqrt{3(x^2-x+1)}}{\sqrt{x^2-x+1}}\right| -\frac{1}{\sqrt{2}}\arctan\left(\frac{\sqrt{2}(1-x)}{\sqrt{x^2+x +1}}\right) + C.$$

JCH
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  • This is the famous technique I once read in Hardy's Pure Mathematics. +1 – Paramanand Singh Sep 08 '17 at 19:38
  • How do you possibly verify this solution? It's very nice mathematics but it's very different than what I get with Mathematica or WolframAlpha (http://www.wolframalpha.com/input/?i=Integrate%5B1%2F((x%5E2+-+x+%2B+1)+Sqrt%5Bx%5E2+%2B+x+%2B+1%5D),+x%5D). This solutions are so long that I can't even tell if they agree with each other. – Cye Waldman Sep 08 '17 at 22:05
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    I have also carried out the solutions numerically and find that your solution does not agree. I do get the same numerical values as with WolframAlpha. – Cye Waldman Sep 08 '17 at 22:29
  • Paranamand Singh. J don't know Hardy's Pure Mathematics. – JCH Sep 09 '17 at 17:45
  • Cye Waldman. Programs Wolfram Alpha and Mathematica are giving very unpleasant result with complex numbers. Y'll attempt to differentiate this result. – JCH Sep 09 '17 at 18:18
3

Substitute $t_{\pm}= \frac{\sqrt{x^2+x+1}}{x\pm 1}$ to establish $$I=\int \frac{x+1}{(x^2-x+1)\sqrt{x^2+x+1}}\ dx =- \int \frac{2}{t_-^2+2} dt_- $$ $$J=\int \frac{x-1}{(x^2-x+1)\sqrt{x^2+x+1}}\ dx = \int \frac{2}{3t_+^2-2} dt_+ $$ Then \begin{align} & \int \frac{1}{(x^2-x+1)\sqrt{x^2+x+1}}\ dx\\ =& \ \frac12(I-J)=- \int \frac1{3t_+^2-2}dt_+-\int \frac1{t_-^2+2}dt_-\\ =& \ \frac1{\sqrt6}\coth^{-1}\frac{\sqrt3 t_+}{\sqrt2} -\frac1{\sqrt2}\tan^{-1}\frac{t_-}{\sqrt2}\\ =& \ \frac1{\sqrt6}\coth^{-1}\frac{\sqrt3 \sqrt{x^2+x+1}}{\sqrt2(x+1)}-\frac1{\sqrt2}\tan^{-1}\frac{\sqrt{x^2+x+1}}{\sqrt2(x-1)} \end{align} As a by-product \begin{align} & \int \frac{x}{(x^2-x+1)\sqrt{x^2+x+1}}\ dx = \frac12(I+J)\\ =& - \frac1{\sqrt6}\coth^{-1}\frac{\sqrt3 \sqrt{x^2+x+1}}{\sqrt2(x+1)}-\frac1{\sqrt2}\tan^{-1}\frac{\sqrt{x^2+x+1}}{\sqrt2(x-1)} \end{align}

Quanto
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