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