int(dx)/((x+1)(x-2))=A log(x+1)+B log(x-2)+C" where "

If int ((x -1) dx )/( (x +1) (x -2))= A log |x +1| +B log |x-2| then A + B =1

int(x-1)/((x-3)(x-2))dx=A ln(x-3)+B ln(x-2)+c , then find the value of A+B

It int (3x-2)/((x-1)(x+2)(x-2))dx=A log |x-1|B log |x+2|+C log |x-3|+C then the ascending order of A,B,C is

int(x dx)/((x-1)(x-2) equal(A) log|((x-1)^2)/(x-2)|+C (B) log|((x-2)^2)/(x-1)|+C (C) log|((x-1)^2)/(x-2)|+C (D) log|(x-1)(x-2)|+C

int(2x-1)/((x-1)(x+2)(x-3))dx=Alog|x-1|+B log|x+2|+C log|x-3|+K Then A,B,C are respectively.

int(x log(x+1)^(2)dx)

int(1)/(x(log x)^(2))dx

int (2x-1)/((x-1)(x+2)(x-3)) dx = A log|x-1|+ B log|x+2|+ C log|x-3|+ K , then A,B,C are respectively