| 1 | 1 | |||
x2 = v | = u' | |||
| 2 | x+1 |
| x2 | x2 − 1 + 1 | 1 | ||||
potem : ∫ | dx = ∫ | dx = ∫(x−1)dx + ∫ | dx | |||
| x+1 | x+1 | x+1 |
| x2 | ||
∫xln(x+1)dx = 0,5*x2*ln(x+1) − 0,5*∫ | dx | |
| x+1 |
| x2 | x2−1 | 1 | ||||
calka ∫ | dx = ∫ | dx + ∫ | dx= | |||
| x+1 | x+1 | x+1 |
| (x−1)(x+1) | ||
=∫ | dx + ln(x+1) = O,5*x2−x + ln(x+1) | |
| x+1 |
| x2 | 1 | 1 | ||||
= xln(x+1) + | − | (x − 1 + | ) = | |||
| 2(x+1) | 2 | x+1 |
| x2 | x2 | |||
= xln(x + 1) + | − | = xln(x+1) ..... zgadza się  | ||
| 2(x+1) | 2(x+1) |
| 1 | ||
v'=x v= | (x2−1) | |
| 2 |
| 1 | ||
u=ln(x+1) du= | ||
| x+1 |