| √1+4x | |
dx | |
| x |
| t2−1 | ||
x= | ||
| 4 |
| 2t2 | ||
wychodzi ∫ | tdt | |
| t2−1 |
| √1+4x | ||
∫ | dx (*) | |
| x |
| t2−1 | ||
t2 = 1+4x ⇒ x = | ||
| 4 |
| 2tdt | |
= 4 | |
| dx |
| tdt | ||
dx = | ||
| 2 |
| t | tdt | 2t2dt | 2(t2−1)+2 | |||||
(*) = ∫ | * | = ∫ | = ∫ | dt = 2t + ∫ | ||||
| (t2−1)/4 | 2 | t2−1 | t2−1 |
| 2 | 1 | 1 | t−1 | |||||
dt = 2t + ∫ ( | − | )dt = 2t + ln(t−1)−ln(t+1) = 2t + ln( | ) = | |||||
| t2−1 | t−1 | t+1 | t+1 |
| √4x+1−1 | ||
2√4x+1+ln( | )+C | |
| √4x+1−1 |