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#