| 1 | ||
2x = t4 − 3 ⇔ x = | (t4 − 3) | |
| 2 |
| 1 | ||
∫x*(2x + 3)14dx = ∫ | (t4 − 3)t*2t3dt = ∫(t4 − 3)t4dt = ∫t16dt − 3∫t4dt = | |
| 2 |
| 1 | 3 | 1 | 3 | |||||
= | t17 − | t5 + C = | (t4)4t − | t4t + C = | ||||
| 17 | 5 | 17 | 5 |
| 1 | 3 | |||
= | (2x + 3)4 * (2x + 3)14 − | ((2x + 3) * (2x + 3)14) + C | ||
| 17 | 5 |
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