Tích phân sau \(\int\limits_0^1 {x\sqrt {{x^2} + 1} } dx = \dfrac{{a\sqrt 2 - b}}{3}\) thì a + b bằng:

* Hướng dẫn giải

Ta có:

\(\begin{array}{l}\int\limits_0^1 {x\sqrt {{x^2} + 1} } dx \\= \dfrac{1}{2}\int\limits_0^1 {\sqrt {{x^2} + 1} \,} d\left( {{x^2} + 1} \right) \\= \dfrac{1}{2}.\dfrac{2}{3}{\left( {{x^2} + 1} \right)^{\dfrac{3}{2}}}\left| \begin{array}{l}^1\\_0\end{array} \right. \\= \dfrac{1}{3}\left( {2\sqrt 2  - 1} \right) = \dfrac{{2\sqrt 2  - 1}}{3}\\\end{array}\)

Khi đó \(\left\{ \begin{array}{l}a = 2\\b = 1\end{array} \right. \Rightarrow a + b = 3.\)

Chọn đáp án C.