Given #dv=x^2sqrt(x^3+1)dx#
Find #v#
#v=int dv=int x^2sqrt(x^3+1)dx#
Integrate by substitution. (There are several notations, I'll choose one.)
#v=int x^2(x^3+1)^(1/2)dx#
Let #w=x^3+1#, so #dw=3x^2 dx#
And our integral becomes:
#v=int x^2(x^3+1)^(1/2)dx=1/3 int (x^3+1)^(1/2)*3x^2dx#
#v=1/3 int (x^3+1)^(1/2)*3x^2dx=1/3 int u^(1/2) du=1/3 2/3u^(3/2) +C#
#v=1/3 2/3u^(3/2) +C=2/9(x^3+1)^(3/2)+C#
A good way of seeing why the #2/9# need to be there is to checl the answer by differentiating:
The derivative of #2/9(x^3+1)^(3/2)# is:
#2/9*[3/2(x^3+1)^(1/2)*3x^2]=(x^3+1)^(1/2)*x^2=x^2sqrt(x^3+1)#
Without the #2/9#, we couldn't cancel #3/2# and #3# to get the original expression.
