| 1 | ||
∫ | dx | |
| x2 * √x2 +1 |
| √tg2(u) + 1 | cos(u) | 1 | ||||
∫ | du = ∫ | du = − | + C = | |||
| tg2(u) | sin2(u) | sin(u) |
| √x2 + 1 | ||
− | + C | |
| x |
| dx | ||
∫ | = | niech x=1t ⇒ dx= − 1t2dt |= | |
| x2√x2+1 |
| − 1t2 dt | − dt | |||
= ∫ | = − ∫ | = | ||
| 1t2√1t2+1 | 1t √1+t2 |
| −tdt | ||
= ∫ | = | i teraz niech √1+t2=z ⇒ 1+t2=z2 ⇒ 2tdt =2zdz |= | |
| √1+t2 |
| −zdz | ||
= ∫ | = −∫dz = − z = − √1+t2 = − √1+ 1x2 = − 1x √x2+1 +C | |
| z |
| 1 | ||
∫ | dx = | x = sh(t)| = | |
| x2 √x2 + 1 |
| 1 | √x2 + 1 | |||
= ∫ | dt = − coth(t) + C = −coth(sh−1(x)) + C = − | + C | ||
| sh2(t) | x |