Giúp mình với ~~~

V

vivietnam

$ x=-t \Longrightarrow dx=-dt$
$I=\int_{-1}^1 \dfrac{e^tdt}{(e^t+1)(t^2+1)}$
$2I=\int_{-1}^1 \dfrac{dx}{x^2+1}=2\int_0^1 \dfrac{dx}{x^2+1}=2arctanx|_0^1=....$
 
You must log in or register to reply here.
Top Bottom