Câu hỏi:

Ta có: $f(x) = \frac{x^2 - 3x + 2}{x} = x - 3 + \frac{2}{x}$.
Vậy, nguyên hàm của $f(x)$ là: $\int f(x) dx = \int (x - 3 + \frac{2}{x}) dx = \frac{x^2}{2} - 3x + 2\ln |x| + C$.

Ta có: $\int\limits_0^9 {\left[ {3f\left( x \right) - 2g\left( x \right)} \right]{\rm{d}}x} = 61$

$\Leftrightarrow 3\int\limits_0^9 {f\left( x \right){\rm{d}}x} - 2\int\limits_0^9 {g\left( x \right){\rm{d}}x} = 61$

$\Leftrightarrow 3.37 - 2\int\limits_0^9 {g\left( x \right){\rm{d}}x} = 61$

$\Leftrightarrow 111 - 2\int\limits_0^9 {g\left( x \right){\rm{d}}x} = 61$

$\Leftrightarrow 2\int\limits_0^9 {g\left( x \right){\rm{d}}x} = 111 - 61 = 50$

$\Leftrightarrow \int\limits_0^9 {g\left( x \right){\rm{d}}x} = -86

Vậy $\int\limits_0^9 {g\left( x \right){\rm{d}}x} = -86$.

Ta có: $\int\limits_0^2 {f\left( x \right){\rm{d}}x} = \int\limits_0^1 {f\left( x \right){\rm{d}}x} + \int\limits_1^2 {f\left( x \right){\rm{d}}x} $
Suy ra: $\int\limits_1^2 {f\left( x \right){\rm{d}}x} = \int\limits_0^2 {f\left( x \right){\rm{d}}x} - \int\limits_0^1 {f\left( x \right){\rm{d}}x} = -4 - 1 = -5$

Ta có: $\int\limits_{ - 1}^4 {f\left( x \right){\rm{d}}x} = - {S_1} + {S_2} = - 7 + 2 = - 5$.
Vậy đáp án là D.

Ta có: $\sin^2 x = \frac{1 - \cos 2x}{2} = \frac{1}{2} - \frac{1}{2} \cos 2x$.
Nguyên hàm của $\sin^2 x$ là: $\int (\frac{1}{2} - \frac{1}{2} \cos 2x) dx = \frac{1}{2}x - \frac{1}{2} . \frac{1}{2} \sin 2x + C = \frac{x}{2} - \frac{\sin 2x}{4} + C$.