" (ii) "(1)/(+1)(x^(2)-x log x+log x-1)/(x-1)

Evaluate the following limits : Lim_(x to 1) (x^(2) - x log x + log x - 1)/(x-1)

2log x-log(x+1)-log(x-1)=

lim_(x rarr1)(x^(3)-x^(2)log x+log x-1)/(x^(2)-1) =

underset( x rarr 1 ) ( "Lim") ( x^(2) - x. ln x + ln x - 1)/( x - 1)

(1)/(x log x log (log x ))

lim_ (x rarr1) (x ^ (2) -x * ln x + ln x-1) / (x-1)

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

int_(1)^(e )(1)/(6x(log x)^(2)+7x log x + 2x)dx=

Evaluate : lim_(x->1)(x^2+x(log)_e x-(log)_e x-1)/((x^2-1)

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