Đáp án:
$B.\, -2$
Giải thích các bước giải:
$\quad \displaystyle\int\limits_0^1\dfrac{x}{x^2 + 2x +1}dx$
$=\displaystyle\int\limits_0^1\left[\dfrac{2x+2}{2(x^2 + 2x +1)}-\dfrac{1}{x^2 + 2x+1}\right]dx$
$= \dfrac12\displaystyle\int\limits_0^1\dfrac{d(x^2 + 2x +1)}{x^2 + 2x +1} - \displaystyle\int\limits_0^1\dfrac{dx}{(x+1)^2}$
$= \dfrac12\ln|x^2 + 2x +1|\Bigg|_0^1 +\dfrac{1}{x+1}\Bigg|_0^1$
$=\dfrac12(\ln4 - \ln1) + \dfrac{1}{1 +1} -\dfrac{1}{0+1}$
$= \dfrac12\ln4 -\dfrac12$
$=-\dfrac12 +\ln2$
$=-\dfrac12 + 1.\ln2 + 0.\ln3$
$\to \begin{cases}a = -\dfrac12\\b = 1\\c = 0\end{cases}$
$\to 6a +b +c = -2$
