How do you integrate #int(3x^5-3)^(3/5)*15x^4 dx# using substitution?

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Answer 1 · 2016-07-24T03:57:40

Let #u = 3x^5 - 3#.

Then #(du)/dx = 15x^4 -> du = 15x^4dx -> (du)/(15x^4) = dx#

#=int(u)^(3/5) xx 15x^4 xx (du)/(15x^4)#

#=int(u^(3/5)) du#

#=(u^(3/5 + 1))/(3/5 + 1) + C#

#= (u^(8/5))/(8/5) + C#

#=5/8u^(8/5) + C#

#=5/8(3x^5 - 3)^(8/5) + C#

Hopefully this helps!