| 1 + x)2 | 1 + 2x + x2 | |||
∫ | *dx = ∫ | *dx = | ||
| x | x |
| 1 | 2x | x2 | ||||
= ∫ | dx + ∫ | dx + ∫ | dx = ln|x| + 2x + x + C | |||
| x | x | x |
| 1 | ||
dobrze mam ? bo w odp mam trochę inny wynik: ln|x| + 2x + | x2 + C | |
| 2 |
| (1+x)2 | 1+2x+x2 | 1 | 2x | x2 | ||||||
∫( | dx=∫ | dx=∫ | dx+∫ | dx+∫ | dx | |||||
| x | x | x | x | x |
| 1 | ||
=∫x−1dx+2∫dx+∫xdx=In|x|+2x+ | x2+C= | |
| 1+1 |
| 1 | ||
=In|x|+2x+ | x2+C | |
| 2 |