1. **State the problem:** We want to find the integral $$\int (\tan x + \sec x)^2 \, dx$$. 2. **Use the formula:** Expand the square using algebraic identity $$(a+b)^2 = a^2 + 2ab + b^2$$: $$ (\tan x + \sec x)^2 = \tan^2 x + 2 \tan x \sec x + \sec^2 x $$ 3. **Rewrite the integral:** $$ \int (\tan^2 x + 2 \tan x \sec x + \sec^2 x) \, dx = \int \tan^2 x \, dx + 2 \int \tan x \sec x \, dx + \int \sec^2 x \, dx $$ 4. **Recall important identities:** - $\frac{d}{dx}(\tan x) = \sec^2 x$ - $\frac{d}{dx}(\sec x) = \sec x \tan x$ - $\tan^2 x = \sec^2 x - 1$ 5. **Evaluate each integral:** - $$\int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx = \int \sec^2 x \, dx - \int 1 \, dx = \tan x - x + C_1$$ - $$2 \int \tan x \sec x \, dx = 2 \int \sec x \tan x \, dx = 2 \sec x + C_2$$ - $$\int \sec^2 x \, dx = \tan x + C_3$$ 6. **Combine all results:** $$ \tan x - x + 2 \sec x + \tan x + C $$ 7. **Simplify:** $$ 2 \tan x + 2 \sec x - x + C $$ **Final answer:** $$ \int (\tan x + \sec x)^2 \, dx = 2 \tan x + 2 \sec x - x + C $$