How to integrate a function of $Ae^{kt} + Be^{-kt}$?

If we integrate by parts, $$\begin{align}\int\frac{(Ae^{kt}-Be^{-kt})^2}{(Ae^{kt}+Be^{-kt})^2}dt&=-\frac1{k(Ae^{kt}+Be^{-kt})}(Ae^{kt}-Be^{-kt})+\frac1k\int\frac{k(Ae^{kt}+Be^{-kt})}{(Ae^{kt}+Be^{-kt})}dt\\ &=-\frac{(Ae^{kt}-Be^{-kt})}{k(Ae^{kt}+Be^{-kt})}+t+C_1\end{align}$$ And $(Ae^{kt}+Be^{-kt})^2-(Ae^{kt}-Be^{-kt})^2=4AB$ so we can say $$\begin{align}I&=\int\frac{C(Ae^{kt}-Be^{-kt})+D}{(Ae^{kt}+Be^{-kt})^2}dt\\ &=\int\frac{C(Ae^{kt}-Be^{-kt})+\frac D{4AB}(Ae^{kt}+Be^{-kt})^2-\frac D{4AB}(Ae^{kt}-Be^{-kt})^2}{(Ae^{kt}+Be^{-kt})^2}dt\\ &=-\frac C{k(Ae^{kt}+Be^{-kt})}+\frac D{4AB}t+\frac D{4AB}\frac{(Ae^{kt}-Be^{-kt})}{k(Ae^{kt}+Be^{-kt})}-\frac D{4AB}t+C_2\\ &=\frac{D(Ae^{kt}-Be^{-kt})-4ABC}{4kAB(Ae^{kt}+Be^{-kt})}+C_2\end{align}$$

Hint Since for any constant $\alpha$ we have $$\frac{d}{dt} e^{\alpha t} = \alpha e^{\alpha t},$$ the derivative of the quantity $$A e^{k t} + B e^{-k t}$$ in parentheses in the denominator is a multiple of the quantity $$A e^{k t} - B e^{-k t}$$ in one of the summands of the numerator, suggesting a natural choice of substitution.

Alternatively, we can rewrite the integral in terms of hyperbolic functions.

Notice that the quantities in parentheses look like hyperbolic functions, albeit with potentially different coefficients of $e^{kt}, e^{-k t}$. By shifting (and, optionally, scaling) the independent variable $t$, we can produce an expression that can be written easily in terms of such functions.

Suppose $k \neq 0$ and that $A, B$ are both positive (or both negative). In terms of the variable $u$ characterized by $$k t = u - u_0$$ we have $$A e^{k t} + B e^{-k t} = A e^{u - u_0} + B e^{-(u - u_0)} = A e^{-u_0} e^u + B e^{u_0} e^{-u} .$$ There is a unique $u_0$ such that $$A e^{-u_0} = B e^{u_0} $$ (we can solve for $u_0$ explicitly, but this turns out not to be necessary), and if for this value of $u_0$ we denote $$\lambda := A e^{-u_0} = B e^{u_0} ,$$ then $$A e^{k t} + B e^{-k t} = \lambda \cosh u \qquad \textrm{and} \qquad A e^{-k t} - B e^{-k t} = \lambda \sinh u .$$

So, we can rewrite the (indefinite) integral as $$\frac{1}{k} \int \frac{C (\lambda \sinh u) + D}{(\lambda \cosh u)^2} du = \underbrace{\frac{C}{k \lambda} \int \tanh u \operatorname{sech} u \,du}_{(1)} + \underbrace{\frac{D}{k \lambda^2} \int \operatorname{sech}^2 u \,du}_{(2)} ,$$ but both of the integrals on the right-hand side are elementary.

For example, the integral in $(1)$ has value $$-\operatorname{sech} u + K = \frac{1}{\cosh u} + K $$ for an arbitrary constant $K$, so term $(1)$ is $$\frac{C}{k \lambda} \left( -\frac{1}{\cosh u} + K \right) = -\frac{C}{k} \cdot \frac{1}{\lambda \cosh u} + K' = -\frac{C}{k (A e^{k t} + B e^{-k t})} + K'$$ for an arbitrary constant $K'$. We can handle term $(2)$ similarly.

We can treat the case that $A$ and $B$ have opposite signs separately, but notice that the antiderivative we produced treating the first case is an antiderivative of the original integrand irrespective of the signs of $A, B$.