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Ta có
$\dfrac{x^2}{x^4-1}=\dfrac{1}{2}.\dfrac{x^2+1+x^2-1}{x^4-1}=\dfrac{1}{2}.\dfrac{1}{x^2-1}+\dfrac{1}{2}.\dfrac{1}{x^2+1}$
$=\dfrac{1}{4}.\dfrac{1}{x-1}+\dfrac{1}{4}.\dfrac{1}{x+1}+\dfrac{1}{2}.\dfrac{1}{x^2+1}$
Suy ra
$\int\limits_{0}^{1/\sqrt 3}\dfrac{x^2}{x^4-1}dx=\left[ {\dfrac{1}{4}\ln|x-1|+\dfrac{1}{4}\ln|x+1| +\dfrac{1}{2}\arctan x} \right]_{0}^{1/\sqrt 3}$
Bạn tự thay nốt kết quả vào nhé.
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