| 2x2 | ||
1) ∫ | dx | |
| x3−17 |
| 1 | ||
t=x3−17 dt=2x2 ∫ | dt = ln |x3−17|+C | |
| t |
| 2 | ||
odpowiedź | ln |x3−17| +C o co chodzi ? | |
| 3 |
| x3 | ||
2) ∫ | dx | |
| sin2(3x4) |
| dt | ||
t=3x4 dt=12x3dx | =x3dx | |
| 12 |
| 1 | 1 | 1 | 1 | |||||
∫ | = | ∫ | co dalej ? | |||||
| 12 | sin2t | 2 | sin2t |
| 1 | ||
odpowiedź ∫− | ctg(3x4)+C | |
| 12 |
| 1 | ||
4) ∫√tg5x * | dx | |
| cos2x |
| e do potęgi 2x | ||
5) ∫ | dx | |
| x2 |
| 1 | ||
6) ∫ | dx | |
| ctgxsin2x |