How do you find the integral #int x^2sqrt(x^3+1)dx# using substitution from [-1,1]?

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Answer 1 · 2016-12-27T15:32:38

The answer is #=4/9sqrt2#

We need

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

We do a substitution

Let #u=x^3+1#, #=>#, #du=3x^2dx#

Therefore,

#int x^2sqrt(x^3+1)dx=1/3intsqrtudu#

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

So,

#int_-1 ^1 x^2sqrt(x^3+1)dx= [2/9(x^3+1)^(3/2)]_ -1^1#

#=4/9sqrt2-0=4/9sqrt2#