Biết \(\int\limits_1^2 {\dfrac{{{x^3}dx}}{{\sqrt {{x^2} + 1} – 1}} = a\sqrt 5 + b\sqrt 2 + c} \) với \(a,b,c\) là các số hữu tỉ. Tính \(P = a + b + c.\)

Đặt \(\sqrt {{x^2} + 1}  = t \Rightarrow {x^2} + 1 = {t^2} \Rightarrow \left\{ \begin{array}{l}xdx = tdt\\{x^2} = {t^2} – 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}dx = \dfrac{t}{x}dt\\{x^2} = {t^2} – 1\end{array} \right.\)

Đổi cận: \(\left\{ \begin{array}{l}x = 1 \Rightarrow t = \sqrt 2 \\t = 2 \Rightarrow t = \sqrt 5 \end{array} \right.\)

Do đó \(\int\limits_1^2 {\dfrac{{{x^3}dx}}{{\sqrt {{x^2} + 1}  – 1}} = \int\limits_{\sqrt 2 }^{\sqrt 5 } {\dfrac{{{x^3}}}{{t – 1}}\dfrac{t}{x}dt = } } \int\limits_{\sqrt 2 }^{\sqrt 5 } {\dfrac{{{x^2}.t}}{{t – 1}}dt = } \int\limits_{\sqrt 2 }^{\sqrt 5 } {\dfrac{{\left( {{t^2} – 1} \right).t}}{{t – 1}}dt = \int\limits_{\sqrt 2 }^{\sqrt 5 } {\dfrac{{\left( {t + 1} \right)\left( {t – 1} \right).t}}{{t – 1}}dt} } \)

\(\begin{array}{l} = \int\limits_{\sqrt 2 }^{\sqrt 5 } {\left( {{t^2} + t} \right)dt = } \left. {\dfrac{{{t^3}}}{3} + \dfrac{{{t^2}}}{2}} \right|_{\sqrt 2 }^{\sqrt 5 } = \dfrac{5}{3}\sqrt 5  + \dfrac{5}{2} – \dfrac{{2\sqrt 2 }}{3} – 1 = \dfrac{5}{3}\sqrt 3  – \dfrac{2}{3}\sqrt 2  + \dfrac{3}{2}\\ \Rightarrow a = \dfrac{5}{3};b =  – \dfrac{2}{3};c = \dfrac{3}{2} \Rightarrow P = a + b + c = \dfrac{5}{2}.\end{array}\)

Chọn C.