Calculus of a Single Variable, Chapter 5, 5.8, Section 5.8, Problem 53

Those 1/x under the hyperbolic cosecant and cotangent are irritating, let's change them to more appropriate y. Make the substitution 1/x = y, then x = 1/y and dx = -1/y^2 dy. The integral becomes
-int (cs ch(y) coth(y))/(1/y^2) (dy)/y^2 = -int cs ch(y) coth(y) dy =
|recall that cs ch(y) = 1/sinh(y) and coth(y) = cosh(y)/sinh(y) |
= -int cosh(y)/(sinh^2(y)) dy.
The next substitution is u = sinh(y), then du = cosh(y) dy, and the integral becomes
- int (du)/u^2 = 1/u + C = 1/sinh(y) + C = 1/sinh(1/x) + C = cs ch(1/x) + C,
where C is an arbitrary constant.