| 1 | ||
∫ | dx | |
| 4x2+2x+1 |
| 2 | 4x+1 | |||
w tym przykładzie wychodzi mi | arctg | , mimo iż w odpowiedziać na początku | ||
| √3 | √3 |
| 1 | ||
jest | ||
| √3 |
| ex−1 | (ex−1)'(ex+1) − (ex−1)(ex+1)' | |||
( | )' = | = | ||
| ex+1 | (ex+1)2 |
| (ex)(ex+1) − (ex−1)ex | |
ile to się będzie równać? | |
| (ex+1)2 |
| (x+√x2+1)n | ||
a. | ||
| n+2 |
| 1 | 1 | |||
b. sin | * | |||
| √n | nx |
| sprezone z | ||
Znajdź liczbe zespoloną z, taką ,ze Re( | =2 oraz Im[(z+1)(1+i)=−3. Ułoz w tym | |
| (1+i) |
| 1 | ||
(3x3−x2−3x+1):(3x+1)=(3x2+x+1)(3x+1)− | ||
| 3 |
| dx | 1 | |||
∫ | , x= | , t≠0 | ||
| x2√1+x2 | t |
| t | ||||||||
= | |||||||||
| dx | dt |
| −t2 | 1 | ||
= | |||
| dx | dt |
| −1 | ||
dx= | *dt | |
| t2 |
| dx | t2dt | t2 | |||||||||||||
∫ | = −∫ | = −∫√ | dt | ||||||||||||
| x2√1+x2 |
| t2+1 |
| 1 | 1 | 2√x | 1 | |||||
∫ | (√x +1)2 dx = ∫ | (x + 2√x +1 ) dx = ∫1 + | + | dx = ∫ | ||||
| x | x | x | x |
| √x | 1 | √x | ||||
dx + 2∫ | dx + ∫ | dx = x + 2∫ | dx + lnx +C = | |||
| x | x | x |
| ∑Xi*Pi | ||
mianowicie, Xc= | ... a więc wpadłem na pomysł, żeby podzielic koło na wiele | |
| P |
| dx | ||
∫ | , x=(t−2)2, t≥2 | |
| 2+√x |
| (t−2)2 | t | ||
= | |||
| dx | dt |
| 2(t−2) | 1 | ||
= | |||
| dx | dt |
| 2(t−2)dt | t−2 | |||
∫ | = 2∫ | dt= | ||
| 2+√(t−2)2 | t |
| t | 2 | t0 | ||||
2∫ | dt−∫ | dt=2t−2∫t−1dt=2t−2 | =2t | |||
| t | t | 0 |
O co tutaj chodzi hehe 
?
Proszę o pomoc
1.