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

Answer 1

Luke Alvarez

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#

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Answer 2

Charles Anctil

To integrate ( \int u^2 \sqrt{u^3 + 2} , du ) by substitution:

Let ( w = u^3 + 2 ), then ( dw = 3u^2 , du ).
Rewrite the integral in terms of ( w ): [ \int u^2 \sqrt{u^3 + 2} , du = \frac{1}{3} \int \sqrt{w} , dw ]
Integrate ( \sqrt{w} ) with respect to ( w ).
Substitute back ( w = u^3 + 2 ) to obtain the final result.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.