20 e2re2xe2ezx
The integral of the function is given as,
∫ e 2 x − e − 2 x e 2 x + e − 2 x d x (1)
Consider, e 2 x + e − 2 x = t .
Differentiate with respect to x .
e 2 x + e − 2 x = t ( 2 e 2 x − 2 e − 2 x ) d x = d t 2 ( e 2 x − e − 2 x ) d x = d t
Substitute 2 ( e 2 x − e − 2 x ) d x = d t in equation (1) and then integrate.
∫ e 2 x − e − 2 x e 2 x + e − 2 x d x = ∫ d t 2 t = 1 2 ∫ 1 t d t = 1 2 log | t | + C
Substitute e 2 x + e − 2 x = t in the above equation.
∫ e 2 x − e − 2 x e 2 x + e − 2 x d x = 1 2 log | t | + C = 1 2 log | e 2 x + e − 2 x | + C
The integral of the function is given as,
Consider,
Differentiate with respect to
Substitute
Substitute
