Đáp án:
$\begin{cases}a = -\dfrac13\\b = -2\\c = 2\end{cases}$
Giải thích các bước giải:
$\quad I =\displaystyle\int\limits_0^3\dfrac{x -1}{1 +\sqrt{1 + x}}$
Đặt $u = \sqrt{1+x}$
$\to du = \dfrac{1}{2\sqrt{1 + x}}dx$
Đổi cận:
$x\quad \Big|\quad 0\qquad 3$
$\overline{u\quad \Big|\quad 1\qquad 2}$
Ta được:
$\quad I = 2\displaystyle\int\limits_1^2\dfrac{u(u^2 -2)}{u+1}du$
$\to I = 2\displaystyle\int\limits_1^2\left(u^2 - u - 1 +\dfrac{1}{u+1}\right)du$
$\to I = 2\displaystyle\int\limits_1^2u^2du - 2\displaystyle\int\limits_1^2udu - 2\displaystyle\int\limits_1^2du + 2\displaystyle\int\limits_1^2\dfrac{1}{u+1}du$
$\to I = \dfrac23u^3\Bigg|_1^2 - u^2\Bigg|_1^2 - 2u\Bigg|_1^2 + 2\ln|u+1|\Bigg|_1^2$
$\to I = \dfrac{14}{3} - 3 - 2+ 2(\ln3- \ln2)$
$\to I = -\dfrac13 - 2\ln2+ 2\ln3$
$\to \begin{cases}a = -\dfrac13\\b = -2\\c = 2\end{cases}$
