[tex]\int_{\frac{1}{12}}^{12}(1+x-\frac{1}{x}).e^{x+\frac{1}{x}}dx[/tex]
[tex]I=\int_{\frac{1}{12}}^{12}(1+x-\frac{1}{x}).e^{x+\frac{1}{x}}dx=\int_{\frac{1}{12}}^{12}e^{x+\frac{1}{x}}dx+\int_{\frac{1}{12}}^{12}(x-\frac{1}{x})e^{x+\frac{1}{x}}dx[/tex]
[tex]I_1=\int_{\frac{1}{12}}^{12}(x-\frac{1}{x})e^{x+\frac{1}{x}}dx=\int_{\frac{1}{12}}^{12}x(1-\frac{1}{x^2})e^{x+\frac{1}{x}}dx[/tex]
sử dụng từng phần cho I1:
[TEX]u=x=>u'=1[/TEX]
[TEX]v'=(1-\frac{1}{x^2})e^{x+\frac{1}{x}}=>v=e^{x+\frac{1}{x}}[/TEX]
=>[tex]I_1=xe^{x+\frac{1}{x}}|_{1/12}^{12}-\int_{\frac{1}{12}}^{12}e^{x+\frac{1}{x}}dx[/tex]
=>I=[tex]xe^{x+\frac{1}{x}}|_{1/12}^{12}[/tex]