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[TEX]\int_{0}^{\sqrt{ln2}}{x^{5}.e^{x^2}dx[/TEX]
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đặt [TEX]x^2=t \Rightarrow 2x.dx=dt [/TEX]
[TEX]\Rightarrow I=\int_{0}^{ln2} \frac{1}{2}.t^2e^tdt=\frac{1}{2} \int_0^{ln2} t^2d(e^t)=\frac{1}{2}.t^2.e^t |_0^{ln2} - \int_0^{ln2}td(e^t) [/TEX]
[TEX] =\frac{1}{2}.t^2.e^t |_0^{ln2} - t.e^t|_0^{ln2} + e^t|_0^{ln2}=............. [/TEX]
 
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