1/2∫2(1+x−x1)ex+x1dx
[TEX]\int_{\frac{1}{2}}^{2}(1+x - \frac{1}{x})e^{x+\frac{1}{x}}[/TEX]
= [TEX]\int_{\frac{1}{2}}^{2}x.(1-\frac{1}{x^2})e^{x+\frac{1}{x}} + \int_{\frac{1}{2}}^{2}e^{x+\frac{1}{x}[/TEX]
= [TEX]x.e^{x+\frac{1}{x}}|_{\frac{1}{2} ^2[/TEX]
ok !
Cái kia là do nếu tớ đặt [TEX]u = e^{x+\frac{1}{x}; v=x[/TEX]
Cái kia có dạng [TEX]u'v+uv'=uv[/TEX]
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