How do you integrate #int x^2sqrt(x^3+1)# using substitution?

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Answer 1 · 2017-06-25T19:39:42

The answer is #=2/9(x^3+1)^(3/2)+C#

We need

#intx^ndx=x^(n+1)/(n+1)+C(n!=-1)#

Perform the substitution

#u=x^3+1#

#du=3x^2dx#

#x^2dx=(du)/3#

Therefore,

#intx^2sqrt(x^3+1)dx=1/3intu^(1/2)du#

#=1/3u^(3/2)/(3/2)#

#=2/9u^(3/2)#

#=2/9(x^3+1)^(3/2)+C#