【正确答案】∫₀¹x²f''(2x)dx=∫₀¹(2x)²f''(2x)d(2x)∫₀²t²f''(t)dt
=∫₀²t²d[f'(t)]=[t²f'(t)｜₀²-2∫₀²tf'(t)dt]
=∫₀²tdf(t)=[tf(t)｜₀²-∫₀²f(t)dt]
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