Ta có
$\int_0^1 x\sqrt{x(x+3)} dx = \int_0^1 x\sqrt{x} \sqrt{x+3} dx$
$= \int_0^1 x^{\frac{3}{2}} \sqrt{x+3} dx$
Ta có
$\dfrac{2}{3} (x^{\frac{3}{2}})' = x^{\frac{1}{2}} = \sqrt{x}$
Vậy
$d(\sqrt{x}) = \dfrac{2}{3} d(x^{\frac{3}{2}})$
Vậy tích phân ban đầu trở thành
$\dfrac{2}{3}\int_0^1 \sqrt{x+3} d(\sqrt{x+3}) = \dfrac{2}{3} \int_{\sqrt{3}}^2 u du$
$= \dfrac{2}{3} \dfrac{u^2}{2}\bigg\vert_{\sqrt{3}}^2$
$= \dfrac{2}{3} (2 - \dfrac{3}{2}) = \dfrac{1}{3}$
Vậy $a = 0, b = \dfrac{1}{3}$
Do đó $4a + 9b = 3$.
