Đáp án:

\( 4x f(x^2) + 3 f(1-x) = \sqrt{1-x^2}. \)

\( \Rightarrow 2 \int_0^1 f(x^2) \, d(x^2) - 3 \int_0^1 f(1-x) \, d(1- x) = \int_0^1 \sqrt{1-x^2} \, dx. \)

\( \Rightarrow 2 \int_0^1 f(x) \, dx - 3 \int_0^1 f(x) \, dx = \int_0^1 \sqrt{1-x^2} \, dx = \frac{\pi}{4}. \)

\( 5I = \frac{\pi}{4} \implies I = \frac{\pi}{20} \Rightarrow \boxed{C}. \)

Đọc nhanh kết quả:

1. \( \int_1^2 f(x) \, dx = 2 \Rightarrow \int_1^4 \frac{f(\sqrt{x})}{\sqrt{x}} \, dx = ? \).

2. \( \int_{-1}^2 f(x) \, dx = 8 \) và \( \int_{-1}^{-3} f(-2x) \, dx = 3 \). Tính \( I = \int_1^6 f(x) \, dx \).

\( 3 = \int_{-1}^{-3} f(-2x) \, dx = -\frac{1}{2} \int_{-1}^{-3} f(-2x) \, d(-2x) = -\frac{1}{2} \int_2^6 f(x) \, dx. \)

\(\Rightarrow \int_2^6 f(x) \, dx = -6 \Rightarrow I = 2-6 =-4\)

Do đó:

\( I = \int_1^6 f(x) \, dx = 2 - 6 = -4. \)