Buồn chém chơi
$I=-\int \dfrac{1}{1-x^2} dx +\int \dfrac{1}{(1-x^2)^2}dx$
Cái $I_1=-\int \dfrac{1}{1-x^2}dx =-\dfrac{1}{2}\int \bigg (\dfrac{1}{1-x}+\dfrac{1}{1+x} \bigg )dx$ dễ nha
Cái $I_2 =\int \dfrac{1}{(1-x^2)^2}dx=\dfrac{1}{4}\int \bigg ( \dfrac{1}{1-x}+\dfrac{1}{1+x}\bigg )^2dx$
$=\dfrac{1}{4}\int \bigg (\dfrac{1}{(1-x)^2} +\dfrac{1}{(1+x)^2}+\dfrac{2}{(1-x)(1+x)} \bigg )$ giống cái $I_1$ khúc cuối tự làm nha