$I_3 = \int \dfrac{4}{x^2 + 2x -3} dx =\int \bigg (\dfrac{1}{x-1} -\dfrac{1}{x+3} \bigg )dx$
$=\ln |x-1| -\ln|x+3| + C =\ln \bigg |\dfrac{x-1}{x+3} \bigg | + C$
$I_4 = \int_{-1}^1 \dfrac{1}{x^2 + 2x +5}dx = \int_{-1}^1 \dfrac{1}{(x+1)^2 +4}dx$
đặt $x+1 =2\tan t \Rightarrow dx = 2\dfrac{dt}{\cos^2 t}$
$I_4 = 2\int \dfrac{1}{\cos^2 t (4\tan^2 t + 4)}dt =\dfrac{1}{2} \int dt =\dfrac{1}{2}t+C$