| 2 + sinx | ||
∫ | dx | |
| 1 + sin2x |
| 1 | sinx | |||
np: = 2∫ | dx + ∫ | dx = .. | ||
| 1 + sin2x | 1 + sin2x |
| dt | dt | |||
... = −∫ | = − ∫ | |||
| 1 + 1 − cos2t | 2 − t2 |
| ln x | ||
∫ | dx | |
| x(4 + ln2x) |
| 1 | ||
podstawiam: ln x=t, | dx = dt | |
| x |
| t | 1 | 1 | ||||
wychodzi: ∫ | dt = ∫ t | dt = tarctgt= | ln xarctgln x | |||
| (4 + t2) | (22 + t2) | 4 |
| 1 | ||
podstawienie: lnx = t, | dx = dt | |
| x |
| 1 | lnx | t | 1 | 2t | ||||||
∫( | * | ) dx = ∫ | dt = | ∫ | dt = | |||||
| x | 4 + ln2x | 4 + t2 | 2 | 4 + t2 |
| 1 | ||
= | ln(4 + t2) + C | |
| 2 |