1. **State the problem:** We want to evaluate the integral $$\int \frac{\cosh x}{\cosh^2 x + \sinh^2 x} \, dx.$$\n\n2. **Simplify the denominator:** Recall the identity $$\cosh^2 x - \sinh^2 x = 1,$$ but here we have $$\cosh^2 x + \sinh^2 x.$$ Using the identity,\n$$\cosh^2 x + \sinh^2 x = (\cosh^2 x - \sinh^2 x) + 2\sinh^2 x = 1 + 2\sinh^2 x.$$\n\n3. **Rewrite the integral:**\n$$\int \frac{\cosh x}{1 + 2\sinh^2 x} \, dx.$$\n\n4. **Use substitution:** Let $$u = \sinh x.$$ Then, $$\frac{du}{dx} = \cosh x$$ which implies $$dx = \frac{du}{\cosh x}.$$\nSince the numerator is $$\cosh x$$, the integral becomes\n$$\int \frac{\cosh x}{1 + 2u^2} dx = \int \frac{\cosh x}{1 + 2u^2} \cdot \frac{du}{\cosh x} = \int \frac{1}{1 + 2u^2} \, du.$$\n\n5. **Integrate the new integral:**\n$$\int \frac{1}{1 + 2u^2} \, du = \int \frac{1}{1 + (\sqrt{2}u)^2} \, du.$$\nRecall the formula $$\int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \arctan\frac{x}{a} + C.$$\nHere, $$a=1$$ and $$x=\sqrt{2}u,$$ so\n$$\int \frac{1}{1 + 2u^2} du = \frac{1}{\sqrt{2}} \arctan (\sqrt{2}u) + C.$$\n\n6. **Replace back the substitution:**\n$$= \frac{1}{\sqrt{2}} \arctan (\sqrt{2} \sinh x) + C.$$\n\n**Final answer:**\n$$\boxed{\int \frac{\cosh x}{\cosh^2 x + \sinh^2 x} \, dx = \frac{1}{\sqrt{2}} \arctan (\sqrt{2} \sinh x) + C}.$$