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 |