Lengthy Indefinite Integral $\int \frac{\sqrt{x^2-4x+3}}{x^2+x+1} dx$

Question

Evaluate the following integral: $$\int \frac{\sqrt{x^2-4x+3}}{x^2+x+1} dx$$

I attempted by multiplying and dividing by the term $\sqrt{x^2-4x+3}$ in the numerator and the denominator and split the numerator $x^2-4x+3$ as follows: $$\frac{\sqrt{x^2-4x+3}}{x^2+x+1}$$ $$=\frac{x^2-4x+3}{\sqrt{x^2-4x+3}(x^2+x+1)}$$ $$=\frac{A(x^2+x+1)+B(x-\omega)+C(x-\omega^2)+D}{\sqrt{x^2-4x+3}(x^2+x+1)}$$ $$$$ where $\omega=e^\frac{2i\pi}{3}$. From here, I got:

$$A=1, B=C=\frac{-5}{2}, D=\frac{9}{2}$$

The integrals involving $A$ and $D$ are straightforward, but I am having problems in solving $B$ and $C$. I tried the substitution $x-\omega^2=\frac{1}{t}$ for the $B$ one and $x-\omega=\frac{1}{t}$ for the $C$ one, but I am getting tedious and lengthy expressions in complex numbers. Is there any shorter way to proceed further or any alternative way to approach this integral?

Answer 1

$$I=\int \frac{\sqrt{x^2-4x+3}}{x^2+x+1}\, dx$$

Using Euler substitution $$\sqrt{x^2-4x+3}=t+x \implies x=\frac{3-t^2}{2 (t+2)}\implies dx=-\frac{t^2+4 t+3}{2 (t+2)^2}$$ $$I=-\int \frac{(t+1)^2 (t+3)^2}{(t+2)\left(t^4-2 t^3-6 t^2+22t+37\right)}\,dt$$

$$t^4-2 t^3-6 t^2+22t+37=(t-a)(t-b)(t-c)(t-d)$$ where

$$(a,b)=\frac{1-i\sqrt{3}}{2}\pm\sqrt{\frac{1}{2}\left(9-5 i \sqrt{3}\right)}$$ $$(c,d)=\frac{1+i\sqrt{3}}{2}\pm\sqrt{\frac{1}{2}\left(9+5 i \sqrt{3}\right)}$$

Using partial fraction decomposition $$\frac{(t+1)^2 (t+3)^2}{(t+2)\left(t^4-2 t^3-6 t^2+22t+37\right)}=$$ $$\frac A{t-a}+\frac B{t-b}+\frac C{t-c}+\frac D{t-d}+\frac E{t+2}$$

Edit

Much simpler is to use @AnuragA suggestion $$x=2+\cosh(t) \implies I=\int \frac{\sinh ^2(t)}{\cosh ^2(t)+5 \cosh(t)+7}\,dt$$ Using the tangent half-angle substitution $$t=2 \tanh ^{-1}(u)\implies I=\int\frac{8 u^2}{(1-u) (u+1) \left(3u^4-12 u^2+13\right)}\,du$$ Partial fraction decomposition $$\frac{8 u^2}{(u-1) (u+1) \left(3u^4-12 u^2+13\right)}=\frac{2 \left(3 u^2-13\right)}{3 u^4-12 u^2+13}-\frac{1}{u-1}+\frac{1}{u+1}$$

For the first term $$\frac{2 \left(3 u^2-13\right)}{3 u^4-12u^2+13}=\frac 2 3 \frac {3 u^2-13}{(u^2-a)(u^2-b)}$$ where $$(a,b)=\frac{6\pm i \sqrt{3}}{3} $$

$$\frac{2 \left(3 u^2-13\right)}{3 u^4-12u^2+13}=\frac{2}{3 (a-b)}\Big(\frac{3 a-13}{u^2-a}-\frac{3 b-13}{u^2-b} \Big)$$ Again, simple integrals.

In other word, this is $x=\frac{u^2-3}{u^2-1}$

Answer 2

Split the integrand into two and, then, apply to both the $3^\text{rd}$ Euler substitution $ t=\frac{\sqrt{(x-3)(x-1)}}{x-1} :$ \begin{align} &\int \frac{\sqrt{x^2-4x+3}}{x^2+x+1}\mathrm dx\\ =& \int \frac1{\sqrt{(x-3)(x-1)}} +\frac{2-5x}{(x^2+x+1) \sqrt{(x-3)(x-1)}}\mathrm dx\\ =&\int \frac2{1-t^2} +\frac{6t^2-26}{3t^4-12t^2+13}\mathrm dt\\ =&\ \ln\bigg|\frac{t+1}{t-1}\bigg| -\frac{\sqrt{13}-\sqrt3}{a_-} \tan^{-1}\frac{\sqrt3t^2-{\sqrt{13}}}{a_- t}\\ &\>\>\>\>\>\>\>\>\>\>\>\>\>\>\> -\frac{\sqrt{13}+\sqrt3}{a_+} \coth^{-1}\frac{\sqrt3t^2+{\sqrt{13}}}{a_+ t}+C \end{align} where $a_\pm = \sqrt{2\sqrt{39}\pm 12}$.

Answer 3

Thanks to @Quanto's answer on my other question, I figured out a really simple way to evaluate this integral. All that needs to be done is substitute $x=t+a, a\in\mathbb R$ such that the constant terms in both the quadratics become the same. In this case, $a=\frac25$. Hence,

$$\begin{align}\int\frac{\sqrt{x^2-4x+3}}{x^2+x+1}\mathrm dx&=\int\frac{\sqrt{t^2-\frac{16t}5+\frac{39}{25}}}{t^2+\frac{9t}5+\frac{39}{25}}\mathrm dt\\&=\text{sgn}(x-2)\cosh^{-1}|x-2|+\frac5{\sqrt{2\sqrt{39}-9}}\tan^{-1}\left(\frac{5x-2+\sqrt{39}}{\sqrt{(2\sqrt{39}-9)(x^2-4x+3)}}\right)-\frac5{\sqrt{2\sqrt{39}+9}}\coth^{-1}\left(\frac{5x-2-\sqrt{39}}{\sqrt{(2\sqrt{39}+9)(x^2-4x+3)}}\right)+C\end{align}$$