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![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⋅](https://static.documents.pub/doc/80x56/60c6e68c0c7c79076f2291a2/subject-mathematics-shift-2-10th-april-2019-2019-4-19-c-a-5-2-d-e-2.jpg)
