t2−1 | ||
t=√2x+1, t2=2x+1, ( | )2=x2 | |
2 |
2 | ||
dt= | dx | |
√2x+1 |
1 | t2 | t2 | ||||||||||||
∫ | dt=2∫ | dt | ||||||||||||
2 |
| (t2−1)2 |
t2 | A | B | C | D | |||||
= | + | + | + | ||||||
(t2−1)2 | t−1 | (t−1)2 | t+1 | (t+1)2 |
1 | t2 | |||
oj, dt= | i masz 4∫ | dt | ||
√2x+1 | (t2−1)2 |
2t | ||
√x= | ||
2−t2 |
4t2 | ||
x= | ||
(2−t2)2 |
t2+2 | ||
√xt+1= | ||
2−t2 |
8t(2−t2)−4t2(−4t) | ||
dx= | dt | |
(2−t2)3 |
8t3+16t | ||
dx= | dt | |
(2−t2)3 |
t2+2 | (2−t2)4 | 8t3+16t | ||
∫ | dt | |||
2−t2 | 16t4 | (2−t2)3 |
1 | (t2+2)2 | ||
∫ | dt | ||
2 | t3 |
1 | t4+4t2+4 | ||
∫ | dt | ||
2 | t3 |
1 | 1 | 8 | |||
(∫2tdt+8∫ | dt+∫ | dt) | |||
4 | t | t3 |
1 | 4 | ||
(t2− | +8ln|t|)+C | ||
4 | t2 |
t4−4 | |
+2ln|t|+C | |
4t2 |
t2−2 | t2+2 | |
+2ln|t|+C | ||
2t | 2t |