| dx | ||
∫ | ||
| x √1 + x5 + x10 |
| −1/t2 | ||
= ∫ | ||
| 1/t √1+1/t5 + 1/t10 |
| t4 | ||
∫ | ||
| √t10 + t5 + 1 |
| dt | ||
Mam 1/5 ∫ | jak to rozgryźć? | |
| t √1 + t + t2 |
| 1 | dt | ||
∫ | |||
| 5 | t√1+t+t2 |
| 1−2u | ||
t = | ||
| u2−1 |
| u−2u2+u2−1 | ||
ut+1= | ||
| u2−1 |
| −(u2−u+1) | ||
ut+1= | ||
| u2−1 |
| −2(u2−1)−2u(1−2u) | ||
dt = | du | |
| (u2−1)2 |
| 2(u2−u+1) | ||
dt = | du | |
| (u2−1)2 |
| 1 | u2−1 | u2−1 | 2(u2−u+1) | ||
∫ | du | ||||
| 5 | 1−2u | −(u2−u+1) | (u2−1)2 |
| 1 | −2 | ||
∫ | du | ||
| 5 | 1−2u |
| 1 | |
ln|1−2u|+C | |
| 5 |
| √1+t+t2−1 | ||
u= | ||
| t |
| 2−2√1+t+t2 | ||
−2u= | ||
| t |
| t+2−2√1+t+t2 | ||
1−2u= | ||
| t |
| 1 | t+2−2√1+t+t2 | ||
ln| | |+C | ||
| 5 | t |