`int(e^(x)-1)/(1-e^(-x))dx=?`
(a) `log(1-e^(x))+c,` (b) `(1)/(e^(x))+c` (c) `e^(x)+c,` (d) `e^(-x)+c`

int(1)/(1+e^(x))dx= (a) log(1+e^(x)) (b) log((1+e^(x))/(e^(x))) (c) log(1+e^(-x)) (d) -log(e^(-x)+1)

int(e^(3x)+e^(5x))/(e^(x)+c^(-x))dx is e q u a l to 4e^(4x)+c(b)e^(4x)+c(e^(4x))/(4)+c(d)2e^(-4x)+c

int_((1)/(e))^(e)|log x|dx=

int x^2 . e^(x^3) dx = (a) e^(x^3) + C (b) e^(x^2) + C (c) 1/3 e^(x^3) + C (d) 1/3 e^(x^2) + C

(d)/(dx){sin^(-1)(e^(x))} is equal to (a) e^(x)sin^(-1)(e^(x)) (b) (e^(x))/(sqrt(1-e^(2x))) (c) (e^(x))/(1-e^(x)) (d) e^(x)cos^(-1)x]]

If int(e^(x)-1)/(e^(x)+1)dx=f(x)+C, then f(x) is equal to

int (e^(x)-e^(-x))/(e^(2x)+e^(-2x))dx=A log|(e^(x)+e^(-x)+a)/(e^(x)+e^(-x)-a)|+c then (A,a) =

int e^(x)(1-cot x+cot^(2)x)dx=e^(x)cot x+C (b) e^(x)cot x+C(c)e^(x)cos ecx+C(d)e^(x)cos ecx+C

int(e^(x)(1+x))/(cos^(2)(xe^(x)))dx=2(log)_(e)cos(xe^(x))+C(b)sec(xe^(x))+C(c)tan(xe^(x))+C(d)tan(x+e^(x))+C

int(1)/(x^(2))e^((x-1)/(x))=......," (A) "(x)/(e^(x-1))+c," (B) "e^((x-1)/(x))+c," (C) "-e^((x)/(x-1))+c," (D) "-e^((x-1)/(x))+c