$I=\int\limits_{-1}^{0} \frac{dx}{x^{2} + 2x + 2} =$$\int\limits_{-1}^{0} \frac{dx}{(x+1)^2 +1}$
Đặt $x+1=\tan t \rightarrow dx = \frac{1}{\cos^2 t}dt = (\tan^2 t+1)dt$
Đổi cận: $x: -1\rightarrow 0$
                $t: 0\rightarrow \frac{\pi }{4}$

$\rightarrow I=\int\limits_{0}^{\frac{\pi }{4}} \frac{(\tan^2 t+1)dt}{\tan^2 t+1}$
           $= \int\limits_{0}^{\frac{\pi }{4}} dt$
           $=\frac{\pi }{4}$

$y= \frac{1}{3}x^3 - x^2  (C)$
$\rightarrow y' = x^2 - 2x$
Gọi $M\in (C) \rightarrow M(a ;\frac{1}{3}a^3-a^2)$
Gọi $\Delta$ là tiếp tuyến của $(C)$ tại M
$\rightarrow \Delta : y=(a^2 -2a).(x-a) +\frac{1}{3}a^3 - a^2$
Hay: $\Delta : y=(a^2-2a)x - \frac{2}{3}a^3+a^2$
*Giả sử $A=\Delta \cap Ox \rightarrow A(\frac{a.(2a-3)}{3.(a-2)};0)  (a\neq 2)$
            $B=\Delta \cap Oy \rightarrow B(0;-\frac{2}{3}a^3+a^2)$
Thao bài ra: $OB=3OA$
                $\Leftrightarrow \sqrt{(\frac{2}{3}a^3-a^2)^2} = 3.\sqrt{(\frac{a.(2a-3)}{3.(a-2)})^2}$
                $\Leftrightarrow \frac{2}{3}a^3-a^2=3.\frac{a.(2a-3)}{3.(a-2)}$
             Hoặc $\frac{2}{3}a^3-a^2=-3.\frac{a.(2a-3)}{3.(a-2)}$
*TH1: $\frac{2}{3}a^3-a^2=3.\frac{a.(2a-3)}{3.(a-2)}$
   $\Leftrightarrow \frac{a^2}{3}.(2a-3) - \frac{a.(2a-3)}{a-2} = 0$
   $\Leftrightarrow (2a-3).(\frac{a^2}{3}-\frac{a}{a-2})=0$
   $\Leftrightarrow (2a-3).(a^3-2a^2-3a)=0$
   $\Leftrightarrow (2a-3).a.(a^2-2a-3)=0$
   $\Leftrightarrow a=\frac{3}{2};                           a=0;                           a=-1;                           a=3$
*TH2: $\frac{2}{3}a^3-a^2=-3.\frac{a.(2a-3)}{3.(a-2)}$
Tương tự có: $a=\frac{3}{2};                       a=0$