How do you integrate #int36x^3(3x^4+3)^5# using substitution?

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Answer 1 · 2016-12-07T10:38:32

The answer is #=729/2(x^4+1)^6+C#

We use #intx^ndx=x^(n+1)/(n+1)*C (n!=-1)#

We do this by substitution

let #u=3(x^4+1)#

#du=3*4x^3dx=12x^3dx#

#x^3dx=(du)/12#

#int36x^3(3x^4+5)^5dx#

#=36/12intu^5du#

#=3u^6/6#

#=1/2u^6=1/2(3^6(x^4+1)^6)+C#

#=729/2(x^4+1)^6+C#