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

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

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

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

int (log x)/(x^(2))dx is equal to a) (log x)/(x) + (1)/(x^(2)) +C b) -(log x)/(x) + (2)/(x) + C c) -(log x)/(x) - (1)/(2x) + C d) -(log x)/(x) - (1)/(x) + C

Find the integral int(1)/(4x^(2)+12x+5)dx A. (1)/(4)log|(2x+1)/(2x-5)|+C B. (1)/(8)log|(1+2x)/(5-2x)|+C C. (1)/(8)log|(2x-1)/(2x+5)|+C D. (1)/(4)log|(1-2x)/(5+2x)|+C

int(log(x+1)-log x)/(x(x+1))dx= (A) log(x-1)log x+(1)/(2)(log x-1)^(2)-(1)/(2)(log x)^(2)+c (B) (1)/(2)(log(x+1))^(2)+(1)/(2)(log x)^(2)-log(x+1)log x+c (C) -(1)/(2)(log(x+1)^(2))-(1)/(2)(log x)^(2)+log x*log(x+1)+c (D) [log(1+(1)/(x))]^(2)+c

int(ln((x-1)/(x+1)))/(x^2-1)dx is equal to (a) 1/2(ln((x-1)/(x+1)))^2+C (b) 1/2(ln((x+1)/(x-1)))^2+C (c) 1/4(ln((x-1)/(x+1)))^2+C (d) 1/4(ln((x+1)/(x-1)))^2+C

int(ln((x-1)/(x+1)))/(x^2-1)dx is equal to (a) 1/2(ln((x-1)/(x+1)))^2+C (b) 1/2(ln((x+1)/(x-1)))^2+C (c) 1/4(ln((x-1)/(x+1)))^2+C (d) 1/4(ln((x+1)/(x-1)))^2+C

int(ln((x-1)/(x+1)))/(x^2-1)dx is equal to (a) 1/2(ln((x-1)/(x+1)))^2+C (b) 1/2(ln((x+1)/(x-1)))^2+C (c) 1/4(ln((x-1)/(x+1)))^2+C (d) 1/4(ln((x+1)/(x-1)))^2+C