[TEX]\int_{-1}^{1}\frac{dx}{(e^x+1)(x^2+1)}=\int_{-1}^{0}\frac{dx}{(e^x+1)(x^2+1)}+\int_{0}^{1}\frac{dx}{(e^x+1)(x^2+1)}=I_1+I_2
I_1=\int_{-1}^{0}\frac{dx}{(e^x+1)(x^2+1)}[/TEX]
đặt [TEX]t=-x => I_1=\int_{0}^{1}\frac{e^tdt}{(e^t+1)(t^2+1)}=\int_{0}^{1}\frac{e^xdx}{(e^x+1)(x^2+1)}
=> I=I_1+I_2=\int_{0}^{1}\frac{e^xdx}{(e^x+1)(x^2+1)}+\int_{0}^{1}\frac{dx}{(e^x+1)(x^2+1)}=\int_{0}^{1}\frac{dx}{(x^2+1)}[/TEX]cái này thì dễ rồi!
[TEX]kq I=\frac{\pi}{4}[/TEX]
[TEX]\int_{-1}^{1}\frac{dx}{(e^x+1)(x^2+1)}=\int_{-1}^{0}\frac{dx}{(e^x+1)(x^2+1)}+\int_{0}^{1}\frac{dx}{(e^x+1)(x^2+1)}=I_1+I_2
I_1=\int_{-1}^{0}\frac{dx}{(e^x+1)(x^2+1)}[/TEX]
đặt [TEX]t=-x => I_1=\int_{0}^{1}\frac{e^tdt}{(e^t+1)(t^2+1)}=\int_{0}^{1}\frac{e^xdx}{(e^x+1)(x^2+1)}
=> I=I_1+I_2=\int_{0}^{1}\frac{e^xdx}{(e^x+1)(x^2+1)}+\int_{0}^{1}\frac{dx}{(e^x+1)(x^2+1)}=\int_{0}^{1}\frac{dx}{(x^2+1)}[/TEX]cái này thì dễ rồi!
[TEX]kq I=\frac{\pi}{4}[/TEX]