You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser.
Giải:
$$\int^1_0 \dfrac{(x^2+2x+2)e^x}{x^2+4x+4}dx$$$$=\int^1_0 \dfrac{(x^2+2x+2)e^x}{(x+2)^2}dx $$$$=-\int^1_0(x^2+2x+2)e^xd\left(\dfrac{1}{x+2}\right) $$$$=-\dfrac{(x^2+2x+2)e^x}{x+2} \bigg|^1_0 + \int^1_0 \dfrac{e^x(2x+2)+e^x(x^2+2x+2)}{x+2}dx $$$$=-\dfrac{(x^2+2x+2)e^x}{x+2} \bigg|^1_0 + \int^1_0 \dfrac{e^x(x^2+4x+4)}{x+2}dx $$$$=-\dfrac{(x^2+2x+2)e^x}{x+2} \bigg|^1_0 + \int^1_0 e^x(x+2)dx $$$$=-\dfrac{(x^2+2x+2)e^x}{x+2} \bigg|^1_0 + \int^1_0 (x+2)de^x $$$$=-\dfrac{(x^2+2x+2)e^x}{x+2} +(x+2)e^x\bigg|^1_0 - \int^1_0 e^xdx $$$$=-\dfrac{(x^2+2x+2)e^x}{x+2} +(x+2)e^x-e^x \bigg|^1_0$$$$=\dfrac{xe^x}{x+2} \bigg|^1_0$$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn