| x2 | ||
∫ | arctg(x) dx | |
| 1 + x2 |
| 1 | |
dx=du | |
| 1+x2 |
| x | ||
∫ arctg x = ∫ x' arctg x dx = x arctg x − ∫ | dx = x arctg x − (1/2) ln (1+x2) | |
| 1+x2 |
| arctg x | ||
nasza ∫ =∫ (arctg x − | ) dx = x arctg x − (1/2) ln (1+x2) − (1/2) (arctg x)2 | |
| 1+x2 |
| x2 +1 | 1 | |||
= ∫ | dx − ∫ | dx ....i już prosto. | ||
| 1 + x2 | 1 + x2 |
| 5^2 | 52 |
| 2^{10} | 210 |
| a_2 | a2 |
| a_{25} | a25 |
| p{2} | √2 |
| p{81} | √81 |
| Kliknij po więcej przykładów | |
|---|---|
| Twój nick | |