How do you use substitution to integrate #(r^2) √(r^(3) +3)dr#?

1 Answer
VNVDVI
Mar 25, 2018

#intr^2/sqrt(r^3+3)dr=2/3sqrt(r^3+3)+C#

Explanation:

So, we want to determine

#intr^2/sqrt(r^3+3)dr#

The substitution should be picked in such a way that the differential of what we've picked will show up in the integral.

The best choice is #u=r^3+3#, as

#du=3r^2dr#

#3r^2dr# does not show up in this exact form in the integral; however, #1/3du=r^2dr# shows up in the numerator of the integral. Thus, we can substitute and now have

#1/3int(du)/sqrt(u)=1/3intu^(-1/2)du# (We've factored #1/3# outside of the integral)

Integrate:

#(1/3)(2)sqrt(u)+C#

Rewriting in terms of #x# yields

#intr^2/sqrt(r^3+3)dr=2/3sqrt(r^3+3)+C#

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