| x−1 | ||
∫ | dx | |
| 3√x+1 |
| (x+1) − 2 | ||
.. = ∫ | dx = ∫(x+1)2/3dx − 2∫(x+1)−1/3dx | |
| 3√x+1 |
| 3 | ||
coś inaczej mi wyszło w odpowiedziach mam ...= | (x−4)3√(x+1)2 | |
| 5 |
| 3 | ||
i tak ma być ... = | (x+1)5/3 − 3(x+1)2/3 =
| |
| 5 |
| 3 | 3 | |||
= | (x+1 − 5)(x+1)2/3 = | (x−4)(x+1)2/3 | ||
| 5 | 5 |
| 3 | 3 | 3 | ||||
= | (x+1)(x+1)2/3 − 3(x+1) = | (x+1 − 5)(x+1)2/3 ... bo | *(−5) = − 3
| |||
| 5 | 5 | 5 |
