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How do you find the integral #int(5x+4)^4dx# using substitution?

1 Answer

Oct 5, 2017

#=1/25(5x+4)^5+C#

Explanation:

By substitution

#int(5x+4)^4dx#

#u=5x+1=>du=5dx=>dx=(du)/5#

#int(5x+4)^4dx=intu^4*(du)/5=dx#

#=1/5intu^4du=1/5xx1/5u^5+C#

#=1/25u^5+C=1/25(5x+4)^5+C#

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