∫ 3x√(x²+1) dx
u=x²+1
u'=2x
2xdx=du =>dx=du/2x
∫ (3x√u)*(du/2x)
∫ (3/2)*√u du
u^(3/2) => (x²+1)^(3/2)
from 4 to 0:
17^(3/2) - 1
u=x²+1
u'=2x
2xdx=du =>dx=du/2x
∫ (3x√u)*(du/2x)
∫ (3/2)*√u du
u^(3/2) => (x²+1)^(3/2)
from 4 to 0:
17^(3/2) - 1
-
Substitution:
u = x² + 1
u = x² + 1