`int(e^(x))/(e^(x)+e^(-x))dx`

inte^(x)(e^(x)-e^(-x))dx=

int_(0)^(1)sqrt((e^(x))/(e^(x)+e^(-x)))dx=

Evaluate: int(e^(x))/(e^(2x)+5e^(x)+6)dx

Evaluate: int(e^(x))/(e^(2x)+6e^(x)+5)dx

Let I =int(e^(x))/(e^(4x)+e^(2x)+1)dx , J = int(e^(-x))/(e^(-4x)+e^(-2x)+1)dx . Then for an arbitary constant C, the value of I - J equals

Statement -1 : If I_(1)=int(e^(x))/(e^(4x)+e^(2x)+1)dx and I_(2)=int(e^(-x))/(e^(-4x)+e^(-2x)+1)dx , then I_(2)-I_(1)=(1)/(2)log((e^(2x)-e^(x)+1)/(e^(2x)+e^(x)+1))+C where C is an arbitrary constant. Statement -2 : A primitive of f(x) =(x^(2)-1)/(x^(4)+x^(2)+1) is (1)/(2)log((x^(2)-x+1)/(x^(2)+x+1)) .

int(e^(x)dx)/(e^(x)-1)

Evaluate : int (e^(x)-e^(-x))/(e^(x)+e^(-x))dx

Evaluate: int(e^(x)-e^(-x))/(e^(x)+e^(-x))dx

Evaluate: int(e^(x)-e^(-x))/(e^(x)+e^(-x))dx