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|làm mệt quá
[TEX]\int_{0}^1 \frac{dx}{(1+x^m)\sqrt[m]{1+x^m}}[/TEX]
[TEX]f(x)=\frac{1}{(1+x^m).\sqrt[m]{(1+x^m)}}dx=\frac{1}{\sqrt[m]{\(1+x^m\)^m}.\sqrt[m]{(1+x^m)}}[/TEX]
[TEX]=\frac{1}{\sqrt[m]{\(1+x^m\)^2}.\sqrt[m]{\(1+x^m\)^{m-2}}.\sqrt[m]{(1+x^m)}} [/TEX]
[TEX]=\frac{1}{\sqrt[m]{\(1+x^m\)^2}.\sqrt[m]{(1+x^m)^{m-1}}} [/TEX]
[TEX]=\frac{1}{ \sqrt[m]{\(1+x^m\)^2}}\[\frac{\(1+x^m\)-x^m}{\sqrt[m]{(1+x^m)^{m-1}}} \][/TEX]
[TEX]=\frac{1}{ \sqrt[m]{\(1+x^m\)^2}}\[\sqrt[m]{1+x^m}-x^m\(1+x^m\)^{\frac{1}{m}-1}\] [/TEX]
[TEX]=\frac{1}{ \sqrt[m]{\(1+x^m\)^2}}\[\sqrt[m]{1+x^m} -\frac{x}{m}\(1+x^m\)^{\frac{1}{m}-1}.m.x^{m-1}\] [/TEX]
[TEX]= \frac{1}{ \sqrt[m]{\(1+x^m\)^2}}\[\ \(x\)'\sqrt[m]{1+x^m}-x.\(\sqrt[m]{1+x^m}\)'\] [/TEX]
[TEX]\Rightarrow F(x)=\frac{x}{\sqrt[m]{1+x^m}}+C[/TEX]
[TEX]\Rightarrow \int_{0}^{1}\frac{1}{(1+x^m).\sqrt[m]{(1+x^m)}}dx=\frac{x}{\sqrt[m]{1+x^m}}\|_{0}^{1}=\frac{1}{\sqrt[m]{2}}[/TEX]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn