Câu hỏi 7

Ta có: \(I = \int\limits_0^1 {\frac{{{x^2} + 2x}}{{{{\left( {x + 3} \right)}^2}}}dx} \) \( = \int\limits_0^1 {\frac{{{x^2} + 2x}}{{{x^2} + 6x + 9}}dx} \) \( = \int\limits_0^1 {\left[ {1 - \frac{{4x + 9}}{{{{\left( {x + 3} \right)}^2}}}} \right]dx} \) \( = \left. x \right|_0^1 - \int\limits_0^1 {\frac{{4x + 9}}{{{{\left( {x + 3} \right)}^2}}}dx} \) \( = 1 - J\)

Với \(J = \int\limits_0^1 {\frac{{4x + 9}}{{{{\left( {x + 3} \right)}^2}}}dx} \) \( = \int\limits_0^1 {\frac{{4x + 12 - 3}}{{{{\left( {x + 3} \right)}^2}}}dx} \) \( = \int\limits_0^1 {\left[ {\frac{4}{{x + 3}} - \frac{3}{{{{\left( {x + 3} \right)}^2}}}} \right]dx} \) \( = 4.\int\limits_0^1 {\frac{{dx}}{{x + 3}}}  - 3.\int\limits_0^1 {\frac{{dx}}{{{{\left( {x + 3} \right)}^2}}}} \)

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

Do đó \(I = 1 - J = 1 - \left( {4\ln \frac{4}{3} - \frac{1}{4}} \right) = \frac{5}{4} - 4\ln \frac{4}{3}\)\( \Rightarrow a = 5,b = 3\) \( \Rightarrow {a^2} + {b^2} = {5^2} + {3^2} = 34\).

Chọn D.