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 $u=3(x^4+1)$

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

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

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

$=36/12intu^5du$

$=3u^6/6$

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

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

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