| x | ||
∫ | ||
| √2+2x2 |
| dt | |
= xdx | |
| 2 |
| 1 | ||
po przekształceniach dochodze do | ∫ (2t+2) (−1/2) dt (2t+2 podnieione do po | |
| 2 |
| −1 | ||
potęgi | ) | |
| 2 |
| 1 | 1 | |||
i potem mi wychodz | * 2√2x2 +2 +C a w odp jest | √2x2 +2 +C zle podstawienie | ||
| 2 | 2 |
| 1 | ||||||||
taka całka to | * (2t+2)1/2 +C co daje 2√2t+2 +C | |||||||
|
| 1 | ||
potem jeszcze | * 2√2t+2 +C = √2t+2 +C | |
| 2 |
| 1 | ||
wykorzystuje ten wzór xndx = | xn+1 | |
| n+1 |
| 1 | 1 | |||
to chyba | * 2 = | tylko dlaczego licze tego pochodna jak mam | ||
| 2√2t+2 | √2t+2 |
| dt | |
= xdx | |
| 4 |
| 1 | 2 | |||
= | ∫t−1/2dt = 2t−1/2 + c = 2*(2+2x2)−1/2 + c = | + C | ||
| 4 | √2+2x2 |
| x | ||
∫ | dx=.. | |
| √2+2x2 |
| 1 | 1 | 1 | 1 | 2 | ||||||
= | ∫ | dt= | ∫t−12dt= | * | t12= | |||||
| 4 | √t | 4 | 4 | 1 |
| 1 | ||
= | √2+2x2+C | |
| 2 |
