| 1 | x+1 | |||
∫ | *ln | dx | ||
| √x | x−1 |
| 1 | ||
∫ln(1+ | ) dx | |
| x |
| 1 | x+1 | x+1 | ||||
∫ | ln | dx = 2 ∫(√x)' ln | dx = | |||
| √x | x−1 | x−1 |
| x+1 | 1 | 1 | ||||
2 √x ln | − 2 ∫ √x ( | − | ) dx = ... | |||
| x−1 | x+1 | x−1 |
| 1 | 1 | |||
Dlaczego | − | |||
| x+1 | x−1 |
| −2 | ||
skoro pochodna ln to jest | , przynajmniej tak wychodzi z moich obliczeń | |
| (x−1)2 |
| x+1 | 1 | 1 | ||||
( ln | )' = (ln (x+1) − ln(x−1) )' = | − | ||||
| x−1 | x+1 | x−1 |