How do you integrate #int(5x^4+5)^(2/3)*20x^3 dx# using substitution?

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Answer 1 · 2016-07-20T14:53:57

Let #u = 5x^4 + 5#, then #(du)/dx = 20x^3 -> du = 20x^3dx -> dx = (du)/(20x^3)#

#=int(u^(2/3)) xx 20x^3 xx (du)/(20x^3)#

#=int(u^(2/3)) xx cancel(20x^3) xx (du)/cancel(20x^3)#

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

#= (u^(5/3))/(5/3) + C#

#= 3/5u^(5/3) + C#

#=3/5(5x^4 + 5)^(5/3) + C#

Hopefully this helps!