How do you find the indefinite integral of #int x^2sqrt(x^3+3)dx#?

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Answer 1 · 2016-12-07T10:22:39

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

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

Lets' do this by substitution

Let #u=x^3+3#

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

Therefore,

#intx^2sqrt(x^3+3)dx#

#=1/3intsqrtudu#

#=1/3intu^(1/2)du#

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

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

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