1. Stating the problem: Evaluate the integral $$\int \frac{1}{1+\cos^\&} \, dx$$. 2. Clarification: The expression $\cos^{\&}$ is unclear. Assuming it is a typo and you meant $\cos^2 x$, the integral becomes $$\int \frac{1}{1+\cos^2 x} \, dx$$. 3. Formula and approach: There is no simple elementary antiderivative for $$\frac{1}{1+\cos^2 x}$$, but we can use the identity $$\cos^2 x = \frac{1+\cos 2x}{2}$$ to rewrite the integral. 4. Substitute the identity: $$ \int \frac{1}{1+\cos^2 x} \, dx = \int \frac{1}{1 + \frac{1+\cos 2x}{2}} \, dx = \int \frac{1}{\frac{2 + 1 + \cos 2x}{2}} \, dx = \int \frac{2}{3 + \cos 2x} \, dx $$ 5. The integral is now $$\int \frac{2}{3 + \cos 2x} \, dx$$. 6. Use the substitution $$t = \tan x$$ and the double angle formula for cosine or use the Weierstrass substitution to solve this integral, but it is a standard integral. 7. The integral $$\int \frac{dx}{a + b \cos x}$$ has the solution $$\frac{2}{\sqrt{a^2 - b^2}} \arctan \left( \frac{(a - b) \tan \frac{x}{2}}{\sqrt{a^2 - b^2}} \right) + C$$ for $$|a| > |b|$$. 8. Here, $$a=3$$ and $$b=1$$, so $$\sqrt{a^2 - b^2} = \sqrt{9 - 1} = \sqrt{8} = 2\sqrt{2}$$. 9. Applying the formula: $$ \int \frac{2}{3 + \cos 2x} \, dx = 2 \times \frac{2}{2\sqrt{2}} \arctan \left( \frac{(3 - 1) \tan x}{2\sqrt{2}} \right) + C = \frac{2}{\sqrt{2}} \arctan \left( \frac{2 \tan x}{2\sqrt{2}} \right) + C $$ 10. Simplify inside the arctan: $$ \frac{2 \tan x}{2\sqrt{2}} = \frac{\tan x}{\sqrt{2}} $$ 11. Final answer: $$ \int \frac{1}{1 + \cos^2 x} \, dx = \frac{2}{\sqrt{2}} \arctan \left( \frac{\tan x}{\sqrt{2}} \right) + C = \sqrt{2} \arctan \left( \frac{\tan x}{\sqrt{2}} \right) + C $$