| 2+√x3−x5 | ||
∫ | ||
| √1−x2 |
| x2+2 | ||
∫ | ||
| 1+x2 |
| 2 + x3/2*√1−x2 | dx | |||
= ∫ | = 2∫ | + ∫x3/2dx = | ||
| √1−x2 | √1−x2 |
| 2 | ||
= 2arcsinx + | x5/2 + c. | |
| 5 |
| x2 + 1 + 1 | 1 | |||
= ∫ | dx = ∫(1 + | )dx = x + arctgx + c. ![]() | ||
| x2 + 1 | x2 + 1 |
| sin2x | ||
∫ | =2cosx dobry wynik mam?![]() | |
| cosx |
| sin2x | 2sinxcosx | |||
∫ | dx = ∫ | dx = 2∫sinxdx = −2cosx + c. ![]() | ||
| cosx | cosx |
bliziutko
| dx | ||
∫ | =tgx ![]() | |
| cos2x + sin2 x |
| 2+32x | ||
∫ | ? | |
| x2(1+x2) |
| x−1 | ||
∫ | ? | |
| √x−1 |