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How do you use substitution to integrate #int(x^4(17-4x^5)^5 dx)#?

1 Answer

Jul 22, 2018

#I=intx^4(17-4x^5)^5dx=int(17-4x^5)^5*x^4dx#
#I=-1/20int[17-4x^5]^5*d(17-4x^5)=-1/20(17-4x^5)^6/6+c#
#:.I=-1/120(17-4x^5)^6+c#

Explanation:

Here ,

#I=intx^4(17-4x^5)^5dx=int(17-4x^5)^5*x^4dx#

Subst. #17-4x^5=u=>-20x^4dx=du=>x^4dx=-1/20du#

So, #I=-1/20intu^5du#

#:,I=-1/20u^6/6+c#

#:.I=-1/120(17-4x^5)^6+c#

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