How do you integrate by substitution $int x^3/sqrt(1+x^4) dx$ ?

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Answer 1 · 2016-12-19T01:52:18

$intx^3/sqrt(1+x^4)dx=sqrt(1+x^4)/2+C$

let $u = 1+x^4 => du = 4x^3dx$

Then, together with the known formula $intx^ndx = x^(n+1)/(n+1)+C$ for $n!=-1$ , we have

$intx^3/sqrt(1+x^4)dx = 1/4int(4x^3)/sqrt(1+x^4)dx$

$=1/4int1/sqrt(u)du$

$=1/4intu^(-1/2)du$

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

$=u^(1/2)/2+C$

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

How do you integrate by substitution int x^3/sqrt(1+x^4) dxint x^3/sqrt(1+x^4) dx?