| e4x−e−x | dt | |||
∫ | dx= |t=ex dt=exdx dx= | | = | ||
| 4 | ex |
| e5x−1 | t5−1 | t5−1 | 1 | t5−1 | ||||||
∫ | dx=∫ | dt=∫ | dt= | ∫ | dt= | |||||
| 4ex | 4t | 4t | 4 | t |
| 1 | 1 | 1 | t5 | |||||
= | ∫(t4−t−1)dt= | (∫t4dt−t−1dt)= | ( | −ln|t|+C)= | ||||
| 4 | 4 | 4 | 5 |
| t5 | ln|t| | e5x | 5ln|ex| | e5x−5x | ||||||
= | − | +C= | − | +C= | +C | |||||
| 20 | 4 | 20 | 20 | 20 |
| e5x+4 | ||
odpowiedź: | +C | |
| 16ex |
| e5x − 1 | dt | |||
= ∫ | = | t = ex , dt = ex dx , dx = | | = | ||
| 4ex | t |
| t5 − 1 | dt | |||
= ∫ | * | = ... | ||
| 4t | t |
| 1 | ||
∫eax dx = | eax + C , a ≠ 0 | |
| a |
