Đáp án:
$T = 12$
Giải thích các bước giải:
$\quad I = \displaystyle\int\limits_0^1\dfrac{3x-1}{x^2 + 6x + 9}dx$
$\Leftrightarrow I = \displaystyle\int\limits_0^1\left[\dfrac{3(2x+6)}{2(x^2 + 6x + 9)} - \dfrac{10}{x^2 + 6x + 9}\right]dx$
$\Leftrightarrow I = \dfrac32\displaystyle\int\limits_0^1\dfrac{2x + 6}{x^2 + 6x + 9}dx - 10\displaystyle\int\limits_0^1\dfrac{1}{x^2 + 6x+ 9}dx$
$\Leftrightarrow I = \dfrac32\displaystyle\int\limits_0^1\dfrac{d(x^2 + 6x + 9)}{x^2 + 6x + 9}- 10\displaystyle\int\limits_0^1\dfrac{d(x+3)}{(x+3)^2}$
$\Leftrightarrow I = \dfrac32\ln(x^2 + 6x + 9)\Bigg|_0^1 + 10\cdot\dfrac{1}{x+3}\Bigg|_0^1$
$\Leftrightarrow I = 3\ln(x+3)\Bigg|_0^1 + 10\cdot\dfrac{1}{x+3}\Bigg|_0^1$
$\Leftrightarrow I = 3(\ln4 - \ln3) + 10\left(\dfrac14 - \dfrac13\right)$
$\Leftrightarrow I = 3\ln\dfrac43 - \dfrac56$
$\Rightarrow \begin{cases}a = 4\\b = 3\end{cases}$
$\Rightarrow ab = 4.3$
$\Rightarrow T = 12$
