| arccosx*dx | ||
∫ | ||
| √x+1 |
| dx | ||
∫ | ||
| 2+√x |
| 4t | |
| 2 + t |
| 2 | ||
√x = x1/2 ⇒ (?)' = x1/2 ? ⇒ ( | x3/2)' = x1/2 | |
| 3 |
| 2 | 2 | 2 | x√x | 1 | ||||||
∫( | x3/2arctg√xdx = | x3/2arctg√x − | ∫ | * | dx | |||||
| 3 | 3 | 3 | x + 1 | 2√x |
| 2 | 1 | x | ||||
= | x3/2arctg√x − | ∫ | dx | | |||
| 3 | 3 | x + 1 |
| 1 | 1 | |||
(√x)' = | , zatem (√x + 1)' = | , przemnóżmy przez 2 i mamy to co | ||
| 2√x | 2√x + 1 |
| 1 | ||
chcemy, mianowicie (2√x + 1)' = | , dalej idzie równie prosto jak 1 | |
| √x + 1 |