How do you integrate by substitution $int u^2sqrt(u^3+2)du$ ?

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Answer 1 · 2016-11-17T06:20:25

The answer is $=2/9(u^3+2)^(3/2) +C$

We use $intv^ndv=v^(n+1)/(n+1)+C (n!=-1)$

Let's do the substitution,

$x=u^3+2$

then, $dx=3u^2du$ $=>$ $u^2du=dx/3$

Therefore, $intu^2sqrt(u^3+2)du=int(sqrtxdx)/3$

$int(x^(1/2)dx)/3=x^(3/2)/(3*3/2)+C$

$=2x^(3/2)/9+C$

$intu^2sqrt(u^3+2)du=2/9(u^3+2)^(3/2) +C$

How do you integrate by substitution int u^2sqrt(u^3+2)duint u^2sqrt(u^3+2)du?