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$I=\int\limits_{0}^{1}\left ( \frac{1}{x^4+x^2+1}+x^3(1-x^2)^3 \right )dx$
$=\int\limits_{0}^{1}\frac{1}{x^4+x^2+1}dx+\int\limits_{0}^{1}x^3(1-x^2)^3dx $
$=I_1+I_2$
Trong đó
$I_2=\int\limits_{0}^{1}x^3(1-x^2)^3dx=\left[ {-\frac{x^{10}}{10}+\frac{3x^{8}}{10}-\frac{x^{6}}{2}+\frac{x^{4}}{4}} \right]_0^1=\frac{1}{40}$
$I_1=\int\limits_{0}^{1}\frac{1}{x^4+x^2+1}dx=\frac{1}{2}\int\limits_{0}^{1}\left ( \frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}\right )dx$
$=\frac{1}{4}\int\limits_{0}^{1}\left ( \frac{d(x^2+x+1)}{x^2+x+1}-\frac{d(x^2-x+1)}{x^2-x+1}\right )+\frac{1}{4}\int\limits_{0}^{1}\left ( \frac{1}{x^2+x+1}+\frac{1}{x^2-x+1}\right )dx$
$=\left[ {\frac{1}{4}\ln\left| {\frac{x^2+x+1}{x^2-x+1}}\right|}+2\sqrt 3\arctan\frac{x-1}{\sqrt 3} +2\sqrt 3\arctan\frac{x-1}{\sqrt 3}\right]_0^1=\frac{\pi \sqrt 3}{12}+\frac{\ln 2}{4}$
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