| x2+1 | ||
∫ | ||
| √3x+1 |
| t2−1 | ||
Podstawiam za √3x+1=t, wyliczam x, wychodzi ze jest on x= | dalej, | |
| 3 |
| 3 | 2√3x+1 | |||
dt= | dx, dx= | dt | ||
| 2√3x+1 | 3 |
| 2t |
| 2 | t2−1 | |||||||||||||
∫ | dt= | ∫ ( | )2+1} dt= | |||||||||||||
| 3 | t | 3 | 3 |
| 2 | 1 | |||
= | * | ∫ (t−1)2+9 dt rozbijając na dwie całki | ||
| 3 | 9 |
| 2 | 2 | 2 | |||
∫(t2−1)2dt= | ∫ ((√3x+1)2−1)2 dx= | ∫ (3x+1−1)2 dx= | |||
| 27 | 27 | 27 |
| 2 | 2 | 2x2 | ||||
= | ∫9x2 dx = | *3x2= | ||||
| 27 | 27 | 9 |
| 2 | 2 | 2 | 2√3x+1 | ||||
∫9 dt= | ∫1 dt= | t = | |||||
| 27 | 3 | 3 | 3 |
| 2√3x+1 | 2x2 | |||
Moje pytanie brzmi, gdzie jest błąd? Licząc pochodną z wyrażenia | + | |||
| 3 | 9 |
| 2t | 1 | |||
[3x+1=t2, 3dx=2tdt, dx= | dt, x= | (t2−1)] | ||
| 3 | 3 |
| x2+1 | 19*(t4−2t2+1)+1 | 2t | ||||
∫ | =∫ | * | dt= | |||
| √3x+1 | t | 3 |
| 2 | ||
= | ∫(t4−2t2+10) dt= | |
| 27 |
| 2 | 1 | 2 | ||||
= | *[ | t5− | t3+10t)= | |||
| 27 | 5 | 3 |
| 2 | 1 | 2 | ||||
= | *[ | √(3x+1)5− | √(3x+1)3+10√3x+1]= | |||
| 27 | 5 | 3 |
| 2 | 1 | 2 | ||||
= | [ | (3x+1)2√3x+1− | *(3x+1)√3x+1+10√3x+1] | |||
| 27 | 5 | 3 |
| 2 | 1 | 2 | ||||
= | √3x+1*[ | *(9x2+6x+1)− | (3x+1)+10]= | |||
| 27 | 5 | 3 |
| 2 | ||
= | √3x+1*[3*(9x2+6x+1)−2*5(3x+1)+10*15]= | |
| 27*15 |