| √x | ||
| | ||
| 1 + √x |
| t*2t | 2t2+2t−2t | 2t*(t+1)−2t | ||||
∫ | dt= ∫ | dt=∫ | = | |||
| 1+t | t+1 | t+1 |
| t | ||
=2∫tdt−2∫ | dt to już dokończysz? | |
| t+1 |
| 1 | ||
[x2=u; 2xdx=du; v'=e−5x; v=∫e−5xdx=− | e−5x] | |
| 5 |
| 1 | 1 | |||
=x2*(− | e−5x)−∫(2x*(− | e−5x)dx= | ||
| 5 | 5 |
| 1 | 2 | |||
=− | x2*e−5x+ | ∫xe−5xdx= jeszcze raz przez części | ||
| 5 | 5 |
| 1 | ||
[x=u; dx=du; v'=e−5x;v=− | e−5x] | |
| 5 |
| 1 | 2 | −1 | 1 | |||||
=− | x2*e−5x+ | *( x* | e−5x+ | ∫e−5xdx)= | ||||
| 5 | 5 | 5 | 5 |