Another tricky

[Tolaso J Kos]

Evaluate the following integral: \( \displaystyle \int_{\phi }^{\frac{1+\sqrt{2}+\sqrt{7+2\sqrt{2}}}{2}}\frac{\left ( x^2+1 \right )\left ( x^2+2x-1 \right )}{x^6+14x^3-1}\, dx \) where \(\phi \) is the golden ratio \( \displaystyle \phi=\frac{1+\sqrt{5}}{2} \).

[Papapetros Vaggelis]

Thank you Tolis for your hint.

Let \(\displaystyle{I=\int_{\phi}^{b}\dfrac{\left(x^2+1\right)\,\left(x^2+2\,x-1\right)}{x^6+14\,x^3-1}\,\rm{dx}}\) ,

where \(\displaystyle{b=\dfrac{1+\sqrt{2}+\sqrt{7+2\,\sqrt{2}}}{2}}\) .

\(\displaystyle{\phi>b>1}\) and

\(\displaystyle{x^6+14\,x^3-1=\left(x^6-1\right)+14\,x^3>0\,,\forall\,x\in\left[\phi,b\right]}\) ,

so \(\displaystyle{I<\infty}\) .

\(\displaystyle{\begin{aligned} I&=\int_{\phi}^{b}\dfrac{\left(x^2+1\right)\,\left(x^2+2\,x-1\right)}{x^6+14\,x^3-1}\,\rm{dx}\\&=\int_{\phi}^{b}\dfrac{\displaystyle{\left(1+\dfrac{1}{x^2}\right)\,\left(x-\dfrac{1}{x}+2\right)}}{\displaystyle{x^3-\dfrac{1}{x^3}+14}}\,\rm{dx}\\&=\int_{\phi}^{b}\dfrac{\displaystyle{\left(1+\dfrac{1}{x^2}\right)\,\left(x-\dfrac{1}{x}+2\right)}}{\displaystyle{\left(x-\dfrac{1}{x}\right)^3+3\,\left(x-\dfrac{1}{x}\right)+14}}\,\rm{dx}\end{aligned}}\)

By substituting \(\displaystyle{u=x-\dfrac{1}{x}\,,u\in\left[1,u_1\right]}\) , where

\(\displaystyle{\begin{aligned}u_1=b-\dfrac{1}{b}&=\dfrac{1+\sqrt{2}+\sqrt{7+2\,\sqrt{2}}}{2}-\dfrac{2}{1+\sqrt{2}+\sqrt{7+2\,\sqrt{2}}}\\&=\dfrac{\left(1+\sqrt{2}+\sqrt{7+2\,\sqrt{2}}\right)^2-4}{2\,\left(1+\sqrt{2}+\sqrt{7+2\,\sqrt{2}}\right)}\\&=\dfrac{\left(1+\sqrt{2}\right)^2+2\,\left(1+\sqrt{2}\right)\,\sqrt{7+2\,\sqrt{2}}+7+2\,\sqrt{2}-4}{2\,\left(1+\sqrt{2}+\sqrt{7+2\,\sqrt{2}}\right)}\\&=\dfrac{\left(1+\sqrt{2}\right)^2+2\,\left(1+\sqrt{2}\right)\,\sqrt{7+2\,\sqrt{2}}+\left(1+\sqrt{2}\right)^2}{2\,\left(1+\sqrt{2}+\sqrt{7+2\,\sqrt{2}}\right)}\\&=\dfrac{2\,\left(1+\sqrt{2}\right)\,\left[\left(1+\sqrt{2}\right)+\sqrt{7+2\,\sqrt{2}}\right]}{2\,\left(1+\sqrt{2}+\sqrt{7+2\,\sqrt{2}}\right)}\\&=1+\sqrt{2}\end{aligned}}\)

we get :

\(\displaystyle{\begin{aligned} I&=\int_{1}^{1+\sqrt{2}}\dfrac{u+2}{u^3+3\,u+14}\,\rm{du}\\&=\int_{1}^{1+\sqrt{2}}\dfrac{u+2}{\left(u+2\right)\,\left(u^2-2\,u+7\right)}\,\rm{du}\\&=\int_{1}^{1+\sqrt{2}}\dfrac{1}{6+\left(u-1\right)^2}\,\rm{du}\\&=\frac{1}{6}\,\int_{1}^{1+\sqrt{2}}\dfrac{1}{\displaystyle{1+\left(\dfrac{u-1}{\sqrt{6}}\right)^2}}\,\rm{du}\\&=\left[\dfrac{1}{\sqrt{6}}\,\arctan\,\left(\dfrac{u-1}{\sqrt{6}}\right)\right]_{1}^{1+\sqrt{2}}\\&=\dfrac{1}{\sqrt{6}}\,\left[\arctan\,\dfrac{1+\sqrt{2}-1}{\sqrt{6}}-\arctan\,\dfrac{1-1}{\sqrt{6}}\right]\\&=\dfrac{1}{\sqrt{6}}\,\arctan\,\dfrac{1}{\sqrt{3}}\\&=\dfrac{\pi}{6\,\sqrt{6}}\end{aligned}}\)

[Tolaso J Kos]

Good evening Vaggelis.
Let me complete the proof to this challenging integral. All that matters is to transform the upper limit, after the change of variables, since the antiderivative has already been calculated by you. Also, the substitution of this integral was pretty straight forward coz' of the lower limit and the equation that it satisfies.

But what happens with the upper limit? Well, after applying the sub \( \displaystyle u=x-\frac{1}{x} \) we have to transform the upper limit somehow. Of course, we cannot plug in that "monster limit" and get a value for \( u \) because this would result in many operations. But, we can do something else.

Let \( b=-(1+\sqrt{2}) \) and square it , thus we get: \( b^2=\left ( 1+\sqrt{2} \right )^2=3+2\sqrt{2} \) which is nice because now we see that \( 7+2\sqrt{2}=3+2\sqrt{2}+4 \) . That whole story reminds us of the quadratic equation , that states :

"If \( ax^2+bx+c=0, \,\,\,a\neq 0 \) then \( x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \)"

Now all the above represent such an equation with \( a=1, \,\,\, b=-(1+\sqrt{2}), \,\,\,\, c=-1\) so the equation is: \(x^2-(1+\sqrt{2})x-1=0 \) and the upper limit is transformed to \( 1+\sqrt{2} \).

Thus, the integral is written as: \( \displaystyle \frac{1}{\sqrt{6}}\tan^{-1}\frac{u-1}{\sqrt{6}}\bigg|_1^{1+\sqrt{2}}=\frac{\sqrt{6}\pi}{36} \)