Đáp án:
$C.\ \dfrac32$
Giải thích các bước giải:
$\quad I = \displaystyle\int\limits_0^3\dfrac{1}{1+ f(x)}dx$
Đặt $x = 3 - t$
$\Rightarrow dx = - dt$
Đổi cận:
$\begin{array}{c|ccc}x&0&&3\\\hline t&3&&0\end{array}$
Ta được:
$\quad I = -\displaystyle\int\limits_3^0\dfrac{1}{1 + f(3 -t)}dt$
$\Leftrightarrow I = \displaystyle\int\limits_0^3\dfrac{1}{1+ f(3 -t)}dt$
$\Leftrightarrow I = \displaystyle\int\limits_0^3\dfrac{1}{1+ f(3 -x)}dx$
$\Leftrightarrow I = \displaystyle\int\limits_0^3\dfrac{1}{1+ \dfrac{1}{f(x)}}dx$
$\Leftrightarrow I = \displaystyle\int\limits_0^3\dfrac{f(x)}{1+ f(x)}dx$
$\Leftrightarrow 2I = \displaystyle\int\limits_0^3\dfrac{1}{1+ f(x)}dx + \displaystyle\int\limits_0^3\dfrac{f(x)}{1+ f(x)}dx$
$\Leftrightarrow 2I = \displaystyle\int\limits_0^3dx$
$\Leftrightarrow 2I = x\Bigg|_0^3$
$\Leftrightarrow 2I = 3$
$\Leftrightarrow I = \dfrac32$
