| x−1 | ||
Całka: ∫ | dx | |
| 3√x+1 |
| x | 1 | |||
wyszło: ∫ | dt − ∫ | dt | ||
| 3√t | 3√t |
| x−1 | x | 1 | ||||
∫ | dx = ∫ | dx − ∫ | dx = I − J // t=x+1 ⇒ dt=dx, x=t−1 | |||
| 3√x+1 | 3√x+1 | 3√x+1 |
| t−1 | t | 1 | t1 | |||||
I= ∫ | dt = ∫ | dt − ∫ | dt = ∫ | dt − | ||||
| 3√t | 3√t | 3√t | t1/3 |
| 33√t2 | 53√t5 | 33√t2 | ||||
∫ t−1/3dt = ∫ t2/3 dt − | + C = | − | + C | |||
| 2 | 3 | 2 |
| 1 | 33√t2 | |||
J= ∫ | dt = ∫ t−1/3 dt = | + C | ||
| 3√t | 2 |
| 53√t5 | 33√t2 | 33√t2 | 53√t5 | |||||
I − J = | − | − | + C = | − 33√t2 + C = | ||||
| 3 | 2 | 2 | 3 |
| 53√(x+1)5 | ||
− 33√(x+1)2 + C mam nadzieje ze sie nie walnalem | ||
| 3 |
u = 3√x+1
u3 = x+1 → x = u3−1.
3u2du = dx.
| x−1 | u3−2 | |||
∫ | dx = ∫ | *3u2du = 3∫(u4−2u)du = ... | ||
| 3√x+1 | u |