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 |
dx | ||
∫ | =tgx | |
cos2x + sin2 x |
2+32x | ||
∫ | ? | |
x2(1+x2) |
x−1 | ||
∫ | ? | |
√x−1 |