Tích phân \(I = \int\limits...

D

- Hướng dẫn giải

Ta có:

\(I = \int\limits_0^1 {\frac{a}{{\sqrt {3{x^2} + 12} }}} dx = \frac{a}{{\sqrt 3 }}\int\limits_0^1 {\frac{1}{{\sqrt {{x^2} + 4} }}} dx\).

Đặt \(u = x + \sqrt {{x^2} + 4} \Rightarrow du = \frac{{x + \sqrt {{x^2} + 4} }}{{\sqrt {{x^2} + 4} }}dx \Rightarrow \frac{{du}}{u} = \frac{{dx}}{{\sqrt {{x^2} + 4} }}\).

\(I = \frac{a}{{\sqrt 3 }}\int\limits_2^{1 + \sqrt 5 } {\frac{1}{u}du} = \left. {\frac{a}{{\sqrt 3 }}\left( {\ln u} \right)} \right|_2^{1 + \sqrt 5 } = \frac{a}{{\sqrt 3 }}\ln \left| {\frac{{1 + \sqrt 5 }}{2}} \right|\).