First, just look at (x-1)/(x+3)^3x−1(x+3)3
(x-1)/(x+3)^3=a/(x+3)+b/(x+3)^2+c/(x+3)^3|* (x+3)^3x−1(x+3)3=ax+3+b(x+3)2+c(x+3)3∣⋅(x+3)3
x-1=a* (x+3)^2+b* (x+3)+cx−1=a⋅(x+3)2+b⋅(x+3)+c
color(red)(0)x^2+color(blue)(1)xcolor(orange)(-1)=(a)x^2+(6a+b)x+9a+3b+c0x2+1x−1=(a)x2+(6a+b)x+9a+3b+c
|(a=color(red)(0)), (6a+b=color(blue)(1)), (9a+3b+c=color(orange)(-1))|
|(a=0), (b=1), (c=-4)|
(x-1)/(x+3)^3=0/(x+3)+1/(x+3)^2-4/(x+3)^3
(x-1)/(x+3)^3=1/(x+3)^2-4/(x+3)^3
int(x-1)/(x+3)^3 dx=int(1/(x+3)^2-4/(x+3)^3)dx
int(1/(x+3)^2-4/(x+3)^3)dx=-1/(x+3)+2/(x+3)^2+c