Đặt $t = \sqrt{x+1}$
Suy ra $t^2 = x + 1$ hay $2tdt = dx$
Thay vào ta có
$\displaystyle \int\limits_1^2 \dfrac{t^2-1}{4 + 2t} 2tdt = \displaystyle \int\limits_1^2 \dfrac{t(t^2-1)}{2 + t} dt$
$= \displaystyle \int\limits_1^2 \dfrac{t^3 - t}{2 + t} dt$
$= \displaystyle \int\limits_1^2 \dfrac{t^3 + 2t^2 - 2t^2 - 4t + 3t + 6 - 6}{t + 2} dt$
$= \displaystyle \int\limits_1^2 \left( t^2 - 2t + 3 - \dfrac{6}{t+2} \right) dt$
$= \left( \dfrac{t^3}{3} - t^2 + 3t - 6\ln|t+2| \right) \Bigg\vert_1^2$
$= \left( \dfrac{8}{3} - 4 + 6 - 6 \ln 4 \right) - \left( \dfrac{1}{3} - 1 + 3 - 6 \ln 3 \right)$
$= \dfrac{7}{3} -12\ln 2 + 6 \ln 3$
Suy ra $a = 7, b = -12, c = 6$. Vậy
$a + b + c = 7 -12 + 6 = 1$
Đáp án A.
