`int(log_e(x+1)-log_ex)/(x(x+1))dx` is equal to (A) `-1/2[log(x+1)^2-1/2logx]^2+log_e(x+1)log_ex+C` (B) `-[(log_e(x+1)-log_ex]^2` (C) `c-1/2(log(1+1/x))^2` (D) none of these

I=int \ log_e (log_ex)/(x(log_e x))dx

int_(1)^(3)|(2-x)log_(e )x|dx is equal to:

int (1+log_e x)/x dx

2log x-log(x+1)-log(x-1) is equals to

int log_e xdx = int 1/(log_x e) dx =

int1/(xcos^2(log_ex))dx

int_(0)^(log 2)(e^(x))/(1+e^(x))dx=

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(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

The integral inte^x(f(x)+f\'(x))dx can be solved by using integration by parts such that: I=inte^xf(x)dx+inte^xf\'(x)dx=e^xf(x)-inte^xf\'(x)dx+inte^xf\'(x)dx=e^xf(x)+C , and inte^(ax)(f(x)+(f\'(x))/a)dx=e^(ax)f(x)/a+C ,Now answer the question: int{log_e(log_ex)+1/(log_ex)^2}dx is equal to (A) log_e(log_ex)+C (B) xlog_e(log_ex)-x/log_ex+C (C) x/log_ex-log_ex+C (D) log_e(log_ex)-x/log_ex+C