Đáp án:

\( \sqrt{f(x) - 1} + e^{f(x)} = \sqrt{x^2 + x + 1} + e^{x^2 + x + 2} \quad (1) \).

Xét \( h(t) = \sqrt{t - 1} + e^t \), hàm này đồng biến trên \( D_h = [1, +\infty) \).

(1) \(\Leftrightarrow h(f(x)) = h(x^2 + x + 2) \Leftrightarrow f(x) = x^2 + x + 2 \).

\(\Rightarrow I = \int_1^2 (x^2 + x + 2) \, dx = \frac{35}{6} \Rightarrow \boxed{C} \).

page51

Đáp án:

\( f(x^5 + 4x + 3) = 2x + 1 \)

\( \Rightarrow (5x^4 + 4)f'(x^5 + 4x + 3) = (5x^4 + 4)(2x + 1) \)

+ \( x^5 + 4x + 3 = -2 \Rightarrow x^5 + 4x + 5 = 0 \quad \text{(chọn \( x = -1 \))} \).

+ \( x^5 + 4x + 3 = 8 \Rightarrow x^5 + 4x - 5 = 0 \quad \text{(chọn \( x = 1 \))} \).

\( \int_{-1}^1 (5x^4 + 4) f(x^5 + 4x + 3) \, dx = \int_{-1}^1 (5x^4 + 4)(2x + 1) \, dx \)

\( = \int_{-2}^8 f(u) \, du = \int_{-1}^1 (5x^4 + 4)(2x + 1) \, dx = 10 \Rightarrow \boxed{B} \)

page52

Đáp án:

Nhắc: \( \int_a^b u'(x) f(u(x)) \, dx = \int_{u(a)}^{u(b)} f(t) \, dt \).

\( \int_0^1 (3x^2 + 3)f(x^3 + 3x + 1) \, dx = \int_0^1 (3x^2 + 3)(3x + 2) \, dx = \frac{59}{4} \)

\( \Rightarrow \int_1^5 f(t) \, dt = \frac{59}{4} \)

\( \int_1^5 x f'(x) \, dx =  x f(x) \Big|_1^5 - \int_1^5 f(x) \, dx \)

\( = 5f(5) - f(1) - \frac{59}{4} \)

Thay \( x = 1 \) vào điều kiện \( \Rightarrow f(5) = 5 \)

Thay \( x = 0 \) vào điều kiện \( \Rightarrow f(1) = 2 \)

\( \int_1^5 x f'(x) \, dx = 25 - 2 - \frac{59}{4} = \frac{33}{4} \Rightarrow \boxed{C} \)

page53

Đáp án:

\( 3 = \int_1^2 f(x-1) \, dx=\int_1^2 f(x-1) \, d(x-1) = \int_0^1 f(t) \, dt \)

\( \int_0^1 x^3 f'(x^2) \, dx = \frac{1}{2} \int_0^1 2x.x^2 f'(x^2) \, dx = \frac{1}{2} \int_0^1 t f'(t) \, dt \)

\( = \frac{1}{2} \left[ t f(t) \Big|_0^1 - \int_0^1 f(t) \, dt \right] = \frac{1}{2} \left[  f(1)  - \int_0^1 f(t) \, dt \right] \)

\( = \frac{1}{2} \left[ 4 - 3 \right] = \frac{1}{2} \Rightarrow \boxed{B}. \)

page54

Đáp án:

\(\int_{-1}^1 g(x)f(x) \, dx = \int_{-1}^1 \big(c_1 - g(- x)\big) f(x) \, dx\)

\(= \int_{-1}^1 f(x) \, dx - \int_{-1}^1 g(- x)f(x) \, dx = 4 + \int_{-1}^1 g(- x) f(- x) \, d(- x)\)

\(\Rightarrow \int_{-1}^1 g(x)f(x) \, dx = 4 + \int_{-1}^1 g(t)f(t) \, dt\)

\(\Rightarrow \int_{-1}^1 g(x)f(x) \, dx = 2 \Rightarrow \boxed{A}\)

page55