a) Sai

Ta có \[f(x) = \frac{1}{{{x^2} - 4}} = \frac{1}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \frac{1}{4}\left( {\frac{1}{{x - 2}} - \frac{1}{{x + 2}}} \right)\]

b) Sai

\[\int\limits_3^4 {f\left( x \right)dx} = \int\limits_3^4 {\frac{1}{{\left( {x - 2} \right)\left( {x + 2} \right)}}dx} = \frac{1}{4}\int\limits_3^4 {\left( {\frac{1}{{x - 2}} - \frac{1}{{x + 2}}} \right)dx} = \frac{1}{4}\left. {\left( {\ln \left( {x - 2} \right) - \ln \left( {x + 2} \right)} \right)} \right|_3^4 = \left. {\frac{1}{4}\ln \frac{{x - 2}}{{x + 2}}} \right|_3^4 = \frac{1}{4}\ln \frac{5}{3}\].

Ta có \[\frac{1}{4}\ln \frac{5}{3} < 0,5\].

c) Đúng

Theo câu b ta có

\[\int\limits_3^4 {f\left( x \right)dx} = \frac{1}{4}\int\limits_3^4 {\left( {\frac{1}{{x - 2}} - \frac{1}{{x + 2}}} \right)dx} = \frac{1}{4}\left. {\left( {\ln \left( {x - 2} \right) - \ln \left( {x + 2} \right)} \right)} \right|_3^4 = \frac{1}{4}\ln \frac{5}{3}\].

Do đó ta có \[a = 5;\,\,b = 3\]. Vậy \[a.b = 15\].

d) Sai

Ta có \[\int\limits_3^4 {\left[ {f\left( x \right) + \frac{{f'\left( x \right)}}{{{f^2}\left( x \right)}}} \right]dx = \int\limits_3^4 {f\left( x \right)} dx + \int\limits_3^4 {\frac{{f'\left( x \right)}}{{{f^2}\left( x \right)}}} dx} \].

Với \[\int\limits_3^4 {f\left( x \right)dx = } \frac{1}{4}\ln \frac{5}{3}\].

\[\int\limits_3^4 {\frac{{f'\left( x \right)}}{{{f^2}\left( x \right)}}} dx = \left. { - \frac{1}{{f\left( x \right)}}} \right|_3^4 = - 12 + 5 = - 7\].

Vậy \[\int\limits_3^4 {\left[ {f\left( x \right) + \frac{{f'\left( x \right)}}{{{f^2}\left( x \right)}}} \right]dx = } \frac{1}{4}\ln \frac{5}{3} - 7\].

Tính tích phân 2x ^3 + 2x ^2 + 1/ x+ 1 dx