[TEX]\int_{}^{}\frac{dx}{x^8+x^6}=\int_{}^{}\frac{(x^2+1-x^2)dx}{x^6(x^2+1)}=\int_{}^{}\frac{dx}{x^6}-\int_{}^{}\frac{dx}{x^4(x^2+1)}[/TEX]
[TEX]=\int_{}^{}\frac{dx}{x^6}-\int_{}^{}\frac{(x^2+1-x^2)dx}{x^4(x^2+1)}=\int_{}^{}\frac{dx}{x^6}-\int_{}^{}\frac{dx}{x^4}+\int_{}^{}\frac{dx}{x^2(x^2+1)}[/TEX]
[TEX]=\int_{}^{}\frac{dx}{x^6}-\int_{}^{}\frac{dx}{x^4}+\int_{}^{}\frac{(x^2+1-x^2)dx}{x^2(x^2+1)}=\int_{}^{}\frac{dx}{x^6}-\int_{}^{}\frac{dx}{x^4}+\int_{}^{}\frac{dx}{x^2}-\int_{}^{}\frac{dx}{x^2+1}[/TEX]
đặt [TEX]x=tant \Rightarrow dx=(1+tan^2t)dt[/TEX]
suy ra[TEX]\frac{dx}{x^2+1}=\int_{}^{}\frac{(1+tan^2t)dt}{1+tan^2t}=\int_{}^{}dt[/TEX]
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