$\dfrac{x^2}{4-x^2}dx=\dfrac{x^2-4}{4-x^2}dx +\dfrac{4}{4-x^2}dx =-dx +\dfrac{4}{4-x^2}dx$
Vậy $I = -\int dx +4\int \dfrac{1}{4-x^2}dx =-x + I_1$
Tính $I_1 = \int \dfrac{1}{4-x^2}dx = \int \dfrac{1}{(2-x)(2+x)}dx =\dfrac{1}{4} \int \bigg (\dfrac{1}{2-x}+\dfrac{1}{2+x} \bigg )dx$
$=\dfrac{1}{4} (\ln |x+2| -\ln |2-x|) +C = \dfrac{1}{4}\ln \bigg |\dfrac{2+x}{2-x} \bigg | +C$
Lắp lên là ok