Report copyright - Subject: Mathematics | Shift 2 | 10th April 2019 · 2019. 4. 19. · C − 5 2 D e 2 Solution Put x2=t 2xdx=dt ∫t2e−t dt 2 = 1 2 [−t2⋅ e−t+2∫te−1dt]+c 1 2 [−t2⋅
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