## a better solution to the last integral

Posted: December 31, 2013 in Mathematics

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

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

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

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

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

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

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

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