[" 6."log|log x^(2)|],[" 8."log|[x+(1)/(x)]|]

int1/(x"log"x^(2))dx is equal to a) 1/(2)"log"|"log"x^(2)|+C b) "log"|"log"x^(2)|+C c) 2"log"|"log"x^(2)|+C d) 4"log"|"log"x^(2)|+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

If "log"_(8){"log"_(2) "log"_(3) (x^(2) -4x +85)} = (1)/(3) , then x equals to

If "log"_(8){"log"_(2) "log"_(3) (x^(2) -4x +85)} = (1)/(3) , then x equals to

Prove that 2 log x-log (x +1)-log (x -1)= 1/x^2+1/(2x^4)+1/(3x^6) +......, where |x|<1.

If log_(4) x + log_(8)x^(2) + log_(16)x^(3) = (23)/(2) , then log_(x) 8 =

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

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

Solve for x:|log^(2)(4-x)+log(4-x)*log(x+(1)/(2))-2log^(2)(x+(1)/(2))=