How do I evaluate #int(15 sqrt(x^3) + 8root3(x^2)) dx#?

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Answer 1 · 2015-02-22T16:04:16

The answer is: #6x^(5/2)+24/5x^(5/3)+c#.

Remembering that:

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

Than:

#int(15x^(3/2)+8x^(2/3))dx=15*x^(3/2+1)/(3/2+1)+8*x^(2/3+1)/(2/3+1)+c=#

#=15*x^(5/2)/(5/2)+8*x^(5/3)/(5/3)+c=15*2/5x^(5/2)+8*3/5x^(5/3)+c=#

#=6x^(5/2)+24/5x^(5/3)+c#.