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Ta có
$ \dfrac{x^4 + 1}{x^6 + 1}= \dfrac{2}{3}.\dfrac{1}{x^2 + 1}+ \dfrac{1}{3}.\dfrac{x^2 + 1}{x^4-x^2 + 1}$
$ \dfrac{x^4 + 1}{x^6 + 1}= \dfrac{2}{3}.\dfrac{1}{x^2 + 1}+ \dfrac{1}{3}.\dfrac{1}{\left ( \dfrac{x}{1-x^2} \right )^2 + 1}.\dfrac{x^2 + 1}{(x^2-1)^2}$
Suy ra
$\int\limits_0^1 {\dfrac{{x^4 + 1}}{{x^6 + 1}}} dx =\int\limits_0^1 \dfrac{2}{3}.\dfrac{1}{x^2 + 1}dx +\int\limits_0^1 \dfrac{1}{3}.\dfrac{1}{\left ( \dfrac{x}{1-x^2} \right )^2 + 1}.d\left ( \dfrac{x}{1-x^2} \right )$
$\int\limits_0^1 {\dfrac{{x^4 + 1}}{{x^6 + 1}}} dx =\left[ { \dfrac{2}{3}\arctan x + \dfrac{1}{3}\arctan\dfrac{x}{1-x^2}} \right]_0^1 $
Đến đây bạn tự thay số vào nhé.
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