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Tính tích phân: [TEX]I=\int_{0}^{1}\frac{dx}{\sqrt[3]{(x^3+1)^4}}[/TEX]
Ta có: [TEX]I=\int_{0}^{1}\frac{dx}{(x^3+1).\sqrt[3]{x^3+1}}[/TEX]
[TEX]I=\int_{0}^{1}\frac{1+x^3-x^3}{(1+x^3)^{\frac{4}{3}}}dx[/TEX]
[TEX]I=\int_{0}^{1}\frac{dx}{(1+x^3)^{\frac{1}{3}}}-\int_{0}^{1}\frac{x^3}{(1+x^3)^{\frac{4}{3}}}dx=I_1+I_2[/TEX]
Tính [TEX]I_2=\int_{0}^{1}\frac{-x^3}{(1+x^3)^{\frac{4}{3}}}dx=\int_{0}^{1}x.\frac{-1}{3}.\frac{3x^2}{(1+x^3)^{\frac{4}{3}}}dx[/TEX]
Đặt [TEX]u=x; dv=\frac{-1}{3}.\frac{3x^2}{(1+x^3)^{\frac{4}{3}}}dx[/TEX]
[TEX]\Rightarrow du=dx; v = \frac{1}{(1+x^3)^{\frac{1}{3}}}[/TEX]
Vậy [TEX]I_2=\frac{x}{(1+x^3)^{\frac{1}{3}}} (can 0; 1) - \int_{0}^{1}\frac{dx}{(1+x^3)^{\frac{1}{3}}}=\frac{x}{(1+x^3)^{\frac{1}{3}}} (can 0; 1) - I_1[/TEX]
Vậy [TEX]I=I_1+I_2=\frac{x}{(1+x^3)^{\frac{1}{3}}} (can 0; 1)=\frac{1}{\sqrt[3]{2}}[/TEX]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
