| 2(t−2) | dt | |||
∫ | dt = 2[ ∫ dt − 2∫ | ] = 2 t − 4 ln t = 2(2+√x) − 4 ln|2 +√x| + C | ||
| t | t |
| dx | ||
∫ | ||
| 3 − √2x +1 |
| (3−t)dt | ||
czyli : ∫ | = 3ln|t| −t = ........ | |
| t |
| xn−1 dx | ||
∫ | , a , b stałe | |
| a + √xn +b |
| 2√xn +b | ||
podstawienie : a + √xn +b = t⇒ √xn +b = t−a⇒dx= | dt | |
| nxn−1 |
| 2(t−a)dt | ||
czyli : dx = | ||
| nxn−1 |
| 2(t−a)dt | 2 | 2 | ||||
∫f(x) dx = ∫ | = | [ ∫ dt − a ∫ dt/t] = | [t −alnt] = ....... | |||
| nt | n | n |