| dx | ||
Oblicz całke ∫ | ||
| 2x2+8x+6 |
| 7 | ||
Rozpisuje to na ułamki i na końcu wychodzi mi | (−ln|x+3|+ln|x+1|) a zamiast 2 powinna | |
| 2 |
| 1 | 1 | |||
= | ∫ | dx | ||
| 2 | (x+1)(x+3) |
| 1 | A | B | |||
= | + | ||||
| (x+1)(x+3) | x+1 | x+3 |
| 1 | 1 | |||
A= | , B=− | |||
| 2 | 2 |
| 1 | 1 | 1 | 1 | 1 | |||||
∫ | − | dx = | ln|x+1|− | ln|x+3|+c | |||||
| 4 | x+1 | x+3 | 4 | 4 |
| 1 | dx | 1 | dx | ||||
∫ | = | ∫ | |||||
| 2 | x2+4x+3 | 2 | (x+1)(x+3) |
| A | B | 1 | |||
+ | = | ||||
| x+1 | x+3 | (x+1)(x+3) |
| 1 | ||
2A = 1 −−> A = | ||
| 2 |
| 1 | ||
−2B = 1 −−> B = − | ||
| 2 |
| 1 | dx | 1 |
|
| |||||||||||||||||||||||
∫ | = | ∫ ( | − | )dx = | |||||||||||||||||||||||
| 2 | (x+1)(x+3) | 2 | x+1 | x+3 |
| 1 | 1 | 1 | 1 | |||||
= | * | ∫ ( | − | ) dx = | ||||
| 2 | 2 | x+1 | x+3 |
| 1 | 1 | |||
= | ln|x+1| − | ln|x+3| | ||
| 4 | 4 |