[Toán 12]Tích phân

N

nguyenbahiep1

[laTEX]u = x^2 \Rightarrow xdx = \frac{1}{2}du \\ \\ I = \frac{1}{2}\int_{0}^{1} \frac{du}{u^2+u+1} \\ \\ I = \frac{1}{2}\int_{0}^{1} \frac{du}{(u+\frac{1}{2})^2 + \frac{3}{4}} \\ \\ u + \frac{1}{2} = \frac{\sqrt{3}}{2}tant[/laTEX]
 
You must log in or register to reply here.
Top Bottom