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

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