| 5 | 1 | 1 | 5 | |||||
−x2−5x−6=−[(x2+5x+6)]=−[(x+ | )2− | ]= | −(x+ | )2 | ||||
| 2 | 4 | 4 | 2 |
| 9x−13 | ||
∫ | dx | |
| √1/4−(x+2,5)2 |
| 5 | 1 | 1 | ||||
[x+ | = | t, dx= | dt] | |||
| 2 | 2 | 2 |
| 1 | ||
x= | t−2,5 i t=2x+5 | |
| 2 |
| 1 |
| |||||||||||
∫ | dt= | |||||||||||
| 2 | √(1/4)−(1/4)t2 |
| 1 | 4,5t−22,5−13 | 4,5t−35,5 | ||||
= | *2∫ | dt= ∫ | dt= | |||
| 2 | √1−t2 | √1−t2 |
| 1 | −2t | 71 | dt | |||||
=4,5*(− | )∫ | dt− | ∫ | dt= skorzystaj | ||||
| 2 | √1−t2 | 2 | √1−t2 |
| u'(x) | ||
[∫ | dx=2u(x) ] | |
| √u(x) |
| −9 | 71 | |||
= | *2√1−t2− | arcsint= | ||
| 4 | 2 |
| −9 | 71 | |||
= | √1−(2x+5)2− | sin(2x+5)= | ||
| 2 | 2 |
| 71 | ||
=−9√−x2−5x−6− | sin(2x+5)+C | |
| 2 |
