Tính các tích phân sau

a) \(\int\limits_0^{\frac{\pi }{3}} {(1 + {{\tan }^2}x)dx} \);               b) \[\int\limits_1^2 {\frac{{{x^2} + 2}}{{2{x^2}}}dx} \];                 c) \(\int\limits_0^1 {\frac{{2{\rm{d}}x}}{{{x^2} + 4x + 3}}} \).

a) \[\int\limits_0^{\frac{\pi }{3}} {(1 + {{\tan }^2}x)dx} = \int\limits_0^{\frac{\pi }{3}} {\frac{1}{{{{\cos }^2}x}}dx} = \left. {\tan x} \right|_0^{\frac{\pi }{3}} = \sqrt 3 - 0 = \sqrt 3 \].

b) \[\int\limits_1^2 {\frac{{{x^2} + 2}}{{2{x^2}}}dx} = \int\limits_1^2 {\left( {\frac{1}{2} + \frac{1}{{{x^2}}}} \right)dx} = \left. {\left( {\frac{1}{2}x - \frac{1}{x}} \right)} \right|_1^2 = 1\]

c) \[\int\limits_0^1 {\frac{{2dx}}{{{x^2} + 4x + 3}}} = \int\limits_0^1 {\frac{2}{{(x + 1)(x + 3)}}} dx = \int\limits_0^1 {\left( {\frac{1}{{x + 1}} - \frac{1}{{x + 3}}} \right)} dx = \left. {\left[ {\ln \left| {x + 1} \right| - \ln \left| {x + 3} \right|} \right]} \right|_0^1 = \ln \frac{3}{2}\].