## a better solution to the last integral

Posted: December 31, 2013 in Mathematics

$\displaystyle\int \ln(x^4+3x^2+1)\;dx$

$\displaystyle=\int\ln\left(x^2+\frac{3-\sqrt5}{2}\right)\;dx+\int\ln\left(x^2+\frac{3+\sqrt5}{2}\right)\;dx$

$\displaystyle=x\ln\left(x^2+\frac{3-\sqrt5}{2}\right)-\int\frac{2x^2}{x^2+\frac{3-\sqrt5}{2}}dx+x\ln\left(x^2+\frac{3+\sqrt5}{2}\right)-\int\frac{2x^2}{x^2+\frac{3+\sqrt5}{2}}dx$

$\displaystyle=x\ln(x^4+3x^2+1)-2\int\frac{x^2}{x^2+\frac{3-\sqrt5}{2}}dx-2\int\frac{x^2}{x^2+\frac{3+\sqrt5}{2}}dx$

$\displaystyle=x\ln(x^4+3x^2+1)-2\int\left(1-\frac{\frac{3-\sqrt5}{2}}{x^2+\frac{3-\sqrt5}{2}}\right)dx-2\int\left(1-\frac{\frac{3+\sqrt5}{2}}{x^2+\frac{3+\sqrt5}{2}}\right)dx$

$=\displaystyle x\ln(x^4+3x^2+1)-4x+2\int\frac{dx}{\frac{2}{3-\sqrt5}x^2+1}+2\int\frac{dx}{\frac{2}{3+\sqrt5}x^2+1}$

$=\displaystyle x\ln(x^4+3x^2+1)-4x+2\sqrt{\frac{3-\sqrt5}{2}}\tan^{-1}\left(\sqrt{\frac{2}{3-\sqrt5}}\;x\right)$

$\displaystyle+2\sqrt{\frac{3+\sqrt5}{2}}\tan^{-1}\left(\sqrt{\frac{2}{3+\sqrt5}}\;x\right)+C$

## relatively “easy” integration

Posted: December 31, 2013 in Mathematics

Due to the holiday season, all my students had gone to vacation so I have a day off playing facebook and saw this integral.  I decided to type this up since I haven’t updated the blog recently.

Solution:

integration by parts,

$\displaystyle\int\ln(x^4+3x^2+1)\;dx$

$\displaystyle=x\ln(x^4+3x^2+1) - \int\frac{4x^4+6x^2}{x^4+3x^2+1}\;dx$

then apply long division onto the integrand,

$\displaystyle=x\ln(x^4+3x^2+1) - \int 4-\frac{6x^2+4}{x^4+3x^2+1}\;dx$

$\displaystyle=x\ln(x^4+3x^2+1) - 4x + \int\frac{6x^2+4}{x^4+3x^2+1}\;dx$

Now, here is the hard part of the whole problem.  We need to integrate this

$\displaystyle \int\frac{6x^2+4}{x^4+3x^2+1}\;dx$

Here we need to break it into two integrals,

$\displaystyle\int\frac{6x^2+6}{x^4+3x^2+1}\;dx-\int\frac{2}{x^4+3x^2+1}\;dx$

$=\displaystyle6\int\frac{x^2+1}{x^4+3x^2+1}\;dx-2\int\frac{dx}{x^4+3x^2+1}$

$=\displaystyle6\int\frac{1+\frac{1}{x^2}}{x^2+3+\frac{1}{x^2}}\;dx-2\int\frac{dx}{\left(x^2+\frac{3-\sqrt5}{2}\right)\left(x^2+\frac{3+\sqrt5}{2}\right)}$

The first integral, let the substitution

$\displaystyle w=x-\frac{1}{x},\quad dw=\left(1+\frac{1}{x^2}\right)\;dx, \quad w^2=x^2+\frac{1}{x^2}-2$

$\displaystyle6\int\frac{1+\frac{1}{x^2}}{x^2+3+\frac{1}{x^2}}\;dx$

$= \displaystyle 6\int\frac{dw}{w^2+5}$

$\displaystyle =\frac{6}{5}\int\frac{dw}{\left(\frac{w}{\sqrt5}\right)^2+1}$

$\displaystyle=\frac{6\sqrt5}{5}\tan^{-1}\frac{w}{\sqrt5}$

$\displaystyle\therefore\int\frac{6x^2+6}{x^4+3x^2+1}dx=\frac{6}{\sqrt5}\tan^{-1}\left(\frac{1}{\sqrt5}\left(x-\frac{1}{x}\right)\right)$

Now the harder one, the second integral,

$\displaystyle-2\int\frac{dx}{\left(x^2+\frac{3-\sqrt5}{2}\right)\left(x^2+\frac{3+\sqrt5}{2}\right)}$

By partial fraction we have,

$\displaystyle\frac{1}{\left(x^2+\frac{3-\sqrt5}{2}\right)\left(x^2+\frac{3+\sqrt5}{2}\right)}=\frac{A}{x^2+\frac{3-\sqrt5}{2}}+\frac{B}{x^2+\frac{3+\sqrt5}{2}}$

$\displaystyle Ax^2+\frac{3+\sqrt5}{2}\cdot A + Bx^2 + \frac{3-\sqrt5}{2}\cdot B = 1$

$\Longrightarrow A+B=0\quad\mbox{and}\quad\dfrac{3+\sqrt5}{2} \cdot A +\dfrac{3-\sqrt5}{2} \cdot B = 1$

$3A+\sqrt5 A+3B-\sqrt5 B=2$

$A-B=\dfrac{2}{\sqrt5}$

$2A=\dfrac{2}{\sqrt5} \Rightarrow A=\dfrac{1}{\sqrt5} \Rightarrow B=-\dfrac{1}{\sqrt5}$

Hence the integral becomes,

$\displaystyle=-2\int\frac{1}{\sqrt5\left(x^2+\frac{3-\sqrt5}{2}\right)}-\frac{1}{\sqrt5\left(x^2+\frac{3+\sqrt5}{2}\right)}\;dx$

$\displaystyle=\frac{2}{\sqrt5}\int\frac{dx}{x^2+\frac{3+\sqrt5}{2}}-\frac{2}{\sqrt5}\int\frac{dx}{x^2+\frac{3-\sqrt5}{2}}$

$=\displaystyle\frac{2}{\sqrt5\left(\frac{3+\sqrt5}{2}\right)}\int\frac{dx}{\left(\sqrt{\frac{2}{3+\sqrt5}}x\right)^2+1}-\frac{2}{\sqrt5\left(\frac{3-\sqrt5}{2}\right)}\int\frac{dx}{\left(\sqrt{\frac{2}{3-\sqrt5}}x\right)^2+1}$

$=\displaystyle\frac{2}{\sqrt5}\sqrt{\frac{2}{3+\sqrt5}}\tan^{-1}\left(\sqrt{\frac{2}{3+\sqrt5}}\;x\right)-\frac{2}{\sqrt5}\sqrt{\frac{2}{3-\sqrt5}}\tan^{-1}\left(\sqrt{\frac{2}{3-\sqrt5}}\;x\right)$

Therefore,

$\displaystyle\int\ln(x^4+3x^2+1)\;dx$

$\displaystyle=x\ln(x^4+3x^2+1)-4x+\frac{6}{\sqrt5}\tan^{-1}\left(\frac{1}{\sqrt5}\left(x-\frac{1}{x}\right)\right)$

$\displaystyle+\frac{2}{\sqrt5}\sqrt{\frac{2}{3+\sqrt5}}\tan^{-1}\left(\sqrt{\frac{2}{3+\sqrt5}}\;x\right)-\frac{2}{\sqrt5}\sqrt{\frac{2}{3-\sqrt5}}\tan^{-1}\left(\sqrt{\frac{2}{3-\sqrt5}}\;x\right)$

$+C$

## solution to the tricky integral

Posted: December 26, 2013 in Mathematics

$\displaystyle\int_{\frac{1+\sqrt5}{2}}^{\frac{1+\sqrt2+\sqrt{7+2\sqrt2}}{2}}\frac{(x^2+1)(x^2+2x-1)}{x^6+14x^3-1}\;dx$

I believe I saw this integral 5 or 6 years ago on AoPS forum.  If my memory serves me right, it was a question from a Japanese university entrance exam.  (I could be wrong, since it’s long time ago… and I am getting the “old people memory disorder” recently…)  The first time when I saw it, I was like WTF?  The integrand already looked hard not to mention the upper limit.  It took me about a week to finally solve it.  (And I had spent a lot of time sitting in the bathroom and thinking hard on it.)  And last month when I post this on my blog, Koopa Koo solved it in a night, or maybe minutes after he saw this.  So this is the difference between a real mathematician and an amateur like me.

Anyway, let’s begin.

Actually the thing that attracted me to solve this integral is the lower limit, it is the famous golden ratio.  I just think that everything that has something to do with the golden ratio is beautiful.  As we all know (if you don’t know then now you know), golden ratio is one of the solution to the equation

$x^2-x-1=0$

and hence,

$x-\dfrac{1}{x}=1$

So if we make a substitution of

$w=x-\dfrac{1}{x}, \quad dw=\left(1+\dfrac{1}{x^2}\right)\;dx$

the lower limit will simplify to 1, what about the upper limit?  Certainly I don’t want to let x equal to that huge expression and then plug it into x-1/x and simplify, that’s not my style coz I am lazy.  When I look at it,

$\dfrac{1+\sqrt2+\sqrt{7+2\sqrt2}}{2}$

It is definitely not the golden ratio although it has a denominator of 2 and some radicals in the numerator and it’s nested too.  How about try to let $b=-(1+\sqrt2)$ and because the expression looks somehow similar to the solution of the quadratic equation.  The quadratic equation states that

$\mbox{If } ax^2+bx+c=0\mbox{ and } a\neq0, \mbox{ then } x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

and let’s try to square $1+\sqrt2$

$(1+\sqrt2)^2=3+2\sqrt2$

This is nice because $7+2\sqrt2 = 3+2\sqrt2+4$ and we can let

$a=1,\quad b=-(1+\sqrt2), \quad c=-1$

hence we have

$x^2-(1+\sqrt2)x-1=0$

$x-\dfrac{1}{x}=1+\sqrt2$

So the upper limit is $1+\sqrt2$

Now the integrand, first we divide top and bottom by $x^3$

$\displaystyle\frac{(x^2+1)(x^2+2x-1)}{x^6+14x^3-1} = \frac{\left(1+\frac{1}{x^2}\right)\left(x+2-\frac{1}{x}\right)}{x^3+14-\frac{1}{x^3}}$

$=\displaystyle\frac{\left(1+\frac{1}{x^2}\right)\left(x+2-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)\left(x^2+1+\frac{1}{x^2}\right)+14}\qquad\mbox{(Difference of cube formula)}$

$=\displaystyle\frac{\left(1+\frac{1}{x^2}\right)\left(x-\frac{1}{x}+2\right)}{\left(x-\frac{1}{x}\right)\left(\left(x-\frac{1}{x}\right)^2+3\right)+14}\qquad\mbox{(Completing the square)}$

then with the following substitution,

$w=x-\dfrac{1}{x}, \quad dw=\left(1+\dfrac{1}{x^2}\right)\;dx$

$\displaystyle\int_{\frac{1+\sqrt5}{2}}^{\frac{1+\sqrt2+\sqrt{7+2\sqrt2}}{2}}\frac{(x^2+1)(x^2+2x-1)}{x^6+14x^3-1}\;dx$

$=\displaystyle\int_1^{1+\sqrt2}\frac{w+2}{w(w^2+3)+14}\;dw$

$=\displaystyle\int_1^{1+\sqrt2}\frac{w+2}{w^3+3w+14}\;dw$

$=\displaystyle\int_1^{1+\sqrt2}\frac{dw}{w^2-2w+7}$

$=\displaystyle\int_1^{1+\sqrt2}\frac{dw}{(w-1)^2+6}$

$=\displaystyle\frac{1}{6}\int_1^{1+\sqrt2}\frac{dw}{\left(\frac{w-1}{\sqrt6}\right)^2+1}$

$=\displaystyle\frac{\sqrt6}{6}\tan^{-1}\frac{w-1}{\sqrt6}\Bigg\vert_1^{1+\sqrt2}$

$=\displaystyle\frac{\sqrt6}{6}\left(\tan^{-1}\frac{1}{\sqrt3}-\tan^{-1}0\right)$

$=\dfrac{\sqrt6\pi}{36}$

🙂