## 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,

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

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

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

$+C$