1. **State the problem:** We need to find the integral $$\int (x \tan x + 1)^2 \, dx$$. 2. **Use the hint and expand the integrand:** $$(x \tan x + 1)^2 = x^2 \tan^2 x + 2x \tan x + 1$$ 3. **Rewrite using the identity:** $$\tan^2 x = \sec^2 x - 1$$ So, $$x^2 \tan^2 x = x^2 (\sec^2 x - 1) = x^2 \sec^2 x - x^2$$ 4. **Rewrite the integral:** $$\int (x \tan x + 1)^2 \, dx = \int (x^2 \sec^2 x - x^2 + 2x \tan x + 1) \, dx$$ 5. **Split the integral:** $$= \int x^2 \sec^2 x \, dx - \int x^2 \, dx + 2 \int x \tan x \, dx + \int 1 \, dx$$ 6. **Integrate the simpler terms:** $$\int x^2 \, dx = \frac{x^3}{3}$$ $$\int 1 \, dx = x$$ 7. **Focus on the integral $$\int x^2 \sec^2 x \, dx$$:** Use integration by parts with $$u = x^2, \quad dv = \sec^2 x \, dx$$ Then, $$du = 2x \, dx, \quad v = \tan x$$ So, $$\int x^2 \sec^2 x \, dx = x^2 \tan x - \int 2x \tan x \, dx$$ 8. **Notice the integral $$\int 2x \tan x \, dx$$ appears again.** Let $$I = \int x \tan x \, dx$$ Use integration by parts again: $$u = x, dv = \tan x \, dx$$ Then, $$du = dx, v = -\ln |\cos x|$$ So, $$I = -x \ln |\cos x| + \int \ln |\cos x| \, dx$$ 9. **Putting it all together:** $$\int x^2 \sec^2 x \, dx = x^2 \tan x - 2I$$ 10. **Rewrite the original integral:** $$\int (x \tan x + 1)^2 \, dx = (x^2 \tan x - 2I) - \frac{x^3}{3} + 2I + x + C$$ 11. **Simplify by canceling $$-2I$$ and $$+2I$$:** $$= x^2 \tan x - \frac{x^3}{3} + x + C$$ **Final answer:** $$\boxed{\int (x \tan x + 1)^2 \, dx = x^2 \tan x - \frac{x^3}{3} + x + C}$$