| x+1 | ||
∫ | dx = | |
| x(x2+1) |
| 1 | x−1 | x−1 | ||||
= ∫ | dx − ∫ | dx = ln|x| − ∫ | dx = ? | |||
| x | x2+1 | x2+1 |
| 2x | ||
∫ | dx = ln(x2+1)+c | |
| x2+1 |
| 1 | ||
∫ | dx = arctg(x)+c | |
| x2+1 |
| x−1 | 1 | 2x | 1 | |||||
∫ | dx= | ∫ | dx−∫ | dx=... | ||||
| x2+1 | 2 | x2+1 | x2+1 |
Ostateczny wynik całki, który otrzymałem to:
| 1 | ||
= ln|x| − | ln|x2+1| + arctgx + C | |
| 2 |