$B = \int\limits_{ - 1}^0 {\frac{{xln(x + 2)}}{{\sqrt {4 - {x^2}} }}} dx$Lưu ý: $y = - \sqrt {4 - {x^2}} \Rightarrow y' = \frac{x}{{\sqrt {4 - {x^2}} }}$
Đặt:
$u = ln(x + 2) \Rightarrow du = \frac{1}{{x + 2}}dx$
$dv = \frac{x}{{\sqrt {4 - {x^2}} }} \Leftarrow v = - \sqrt {4 - {x^2}} $
$B = - \sqrt {4 - {x^2}} ln(x + 2)\begin{cases}0 \\ -1 \end{cases} + \int\limits_{ - 1}^0 {\frac{{\sqrt {4 - {x^2}} }}{{x + 2}}} dx = - 2\ln 2 + \int\limits_{ - 1}^0 {\frac{{\sqrt {4 - {x^2}} }}{{x + 2}}} dx$
Đặt $x = 2\sin t,t \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$
$dx = 2\cos tdt$
$B = - 2\ln 2 + \int\limits_{ - \frac{\pi }{6}}^0 {\frac{{\sqrt {4 - 4{{\sin }^2}t} }}{{2\sin t + 2}}.2} \cos tdt = - 2\ln 2 + \int\limits_{ - \frac{\pi }{6}}^0 {\frac{{2{{\cos }^2}t}}{{\sin t + 1}}} dt$
$= - 2\ln 2 + 2\int\limits_{ - \frac{\pi }{6}}^0 {(1 - \sin t)} dt = - 2\ln 2 + 2 - \sqrt 3 + \frac{\pi }{3}$