1. **Problem statement:** Evaluate the integral $$\int_{\pi/6}^{\pi/3} \frac{3\tan^2 x + 6 \tan x + 11}{1 + \tan^2 x} \, dx = \frac{\pi a + b}{6}$$ and find $a+b$. 2. **Recall the identity:** $$1 + \tan^2 x = \sec^2 x$$ So the integrand can be rewritten as $$\frac{3\tan^2 x + 6 \tan x + 11}{1 + \tan^2 x} = \frac{3\tan^2 x + 6 \tan x + 11}{\sec^2 x} = (3\tan^2 x + 6 \tan x + 11) \cos^2 x$$ 3. **Express $\tan x$ and $\tan^2 x$ in terms of $\sin x$ and $\cos x$: ** $$\tan x = \frac{\sin x}{\cos x}, \quad \tan^2 x = \frac{\sin^2 x}{\cos^2 x}$$ Substitute into the integrand: $$ (3 \frac{\sin^2 x}{\cos^2 x} + 6 \frac{\sin x}{\cos x} + 11) \cos^2 x = 3 \sin^2 x + 6 \sin x \cos x + 11 \cos^2 x$$ 4. **Simplify the integrand:** $$3 \sin^2 x + 6 \sin x \cos x + 11 \cos^2 x$$ 5. **Use the Pythagorean identity:** $$\sin^2 x + \cos^2 x = 1$$ Rewrite $3 \sin^2 x + 11 \cos^2 x$ as $$3 \sin^2 x + 11 \cos^2 x = 3(1 - \cos^2 x) + 11 \cos^2 x = 3 + 8 \cos^2 x$$ So integrand becomes $$3 + 8 \cos^2 x + 6 \sin x \cos x$$ 6. **Rewrite $\cos^2 x$ using double angle formula:** $$\cos^2 x = \frac{1 + \cos 2x}{2}$$ So $$8 \cos^2 x = 8 \times \frac{1 + \cos 2x}{2} = 4 + 4 \cos 2x$$ 7. **Substitute back:** Integrand = $$3 + 4 + 4 \cos 2x + 6 \sin x \cos x = 7 + 4 \cos 2x + 6 \sin x \cos x$$ 8. **Rewrite $6 \sin x \cos x$ using double angle formula:** $$\sin 2x = 2 \sin x \cos x \implies 6 \sin x \cos x = 3 \sin 2x$$ 9. **Final integrand:** $$7 + 4 \cos 2x + 3 \sin 2x$$ 10. **Integrate term by term:** $$\int_{\pi/6}^{\pi/3} 7 \, dx = 7 \left( \frac{\pi}{3} - \frac{\pi}{6} \right) = 7 \times \frac{\pi}{6} = \frac{7\pi}{6}$$ $$\int_{\pi/6}^{\pi/3} 4 \cos 2x \, dx = 4 \times \frac{\sin 2x}{2} \Big|_{\pi/6}^{\pi/3} = 2 (\sin \frac{2\pi}{3} - \sin \frac{\pi}{3})$$ Calculate sines: $$\sin \frac{2\pi}{3} = \sin 120^\circ = \frac{\sqrt{3}}{2}, \quad \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$$ So difference is zero, integral is zero. 11. **Integrate $3 \sin 2x$:** $$\int_{\pi/6}^{\pi/3} 3 \sin 2x \, dx = 3 \times \left(-\frac{\cos 2x}{2}\right) \Big|_{\pi/6}^{\pi/3} = -\frac{3}{2} (\cos \frac{2\pi}{3} - \cos \frac{\pi}{3})$$ Calculate cosines: $$\cos \frac{2\pi}{3} = \cos 120^\circ = -\frac{1}{2}, \quad \cos \frac{\pi}{3} = \frac{1}{2}$$ So $$-\frac{3}{2} \left(-\frac{1}{2} - \frac{1}{2}\right) = -\frac{3}{2} (-1) = \frac{3}{2}$$ 12. **Sum all integrals:** $$\frac{7\pi}{6} + 0 + \frac{3}{2} = \frac{7\pi}{6} + \frac{3}{2}$$ Rewrite $\frac{3}{2}$ as $\frac{9}{6}$ to have common denominator 6: $$\frac{7\pi}{6} + \frac{9}{6} = \frac{7\pi + 9}{6}$$ 13. **Compare with given form:** $$\frac{\pi a + b}{6} = \frac{7\pi + 9}{6}$$ So $$a = 7, \quad b = 9$$ Therefore, $$a + b = 7 + 9 = 16$$ **Final answer:** 16