phân tích $\dfrac{x^2}{(x+1)^4}dx = \dfrac{(x+1)^2}{(x+1)^4}dx -\dfrac{2x+1}{(x+1)^4}dx = \dfrac{1}{(x+1)^2}dx - \dfrac{2(x+1)}{(x+1)^4}dx +\dfrac{1}{(x+1)^4}dx$
$=\dfrac{1}{(x+1)^2}dx - \dfrac{2}{(x+1)^3}dx +\dfrac{1}{(x+1)^4}dx = \dfrac{d(x+1)}{(x+1)^2} - \dfrac{2d(x+1)}{(x+1)^3}+\dfrac{d(x+1)}{(x+1)^4}$
toàn cái dễ rồi nhé
Cau 2 phân tích
$I=\dfrac{1-e^x}{1+e^x}dx =\dfrac{(1-e^x)e^x dx}{e^x (1+e^x)}dx$ đặt $e^x +1 = t \Rightarrow e^x dx =dt$
$I=\dfrac{1-(t-1)}{(t-1)t}dt=\dfrac{2-t}{t(t-1)}dt =\dfrac{2}{t(t-1)}dt -\dfrac{1}{t-1}dt$
$=2 \bigg (\dfrac{1}{t-1} -\dfrac{1}{t} \bigg )dt -\dfrac{d(t-1)}{t-1}$ Dễ rồi nhé