Calki
koligdedb: Musze podstawic
∫x√1+xdx
8 lis 22:17
Rafał28:
t = √x + 1
8 lis 22:30
pigor: ..., np. tak :
∫x√1+x dx= |
√1+x= t ⇒ 1+x=t
2 i x=t
2−1 i dx=2tdt | =
= ∫(t
2−1)*t*2t dt= 2 ∫ (t
4−t
2) dt= 2 (
15t
5−
13t
3) dt=
215t
3 (3t
2−5)=
=
215 (1+x)
√1+x (3(1+x)−5)=
215(1+x)(3x−2)√1+x+C . ...
8 lis 22:34
ICSP: = ∫(x+1)
√x+1 dx − ∫
√x+1 = ∫(x+1)
3/2 dx − ∫
√x+1 dx =
| | 2 | | 2 | |
= |
| (x + 1)5/2 − |
| (x+1)3/2 + C |
| | 5 | | 3 | |
8 lis 22:57
nuggets:
ubek ∨ patriota
?
9 lis 02:13