| Dave's Math Tables:
Integral tanh(x)  | 
| (Math | Calculus | Integrals | Table Of | tanh x) | 
|  tanh x dx = ln (cosh x) + C. | 
|  | sinh x cosh x | = | (ex + e-x) / 2 | 
|  tanh
x dx =  | ex + e-x | dx | 
set
  u = ex + e-x
then we find
  du = (ex - e-x) dx
substitute du= (ex - e-x) dx, u = ex
+ e-x
 
| =  | u | 
solve
= ln |u| + C
substitute back u = ex + e-x
= ln |ex + e-x| + C
since ex and e-x are always positive
= ln (ex + e-x) + C
since (ex + e-x)/2 = cosh(x)
= ln (2 cosh x) + C
= ln 2 + ln (cosh x) + C
ln 2 is merely a constant that can be combined with C
= ln (cosh x) + C
Q.E.D.
