How do you use substitution to integrate $(r^{2}) \sqrt (r^{3} + 3) d r$ $(r^2) √(r^(3) +3)dr$ ?

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How do you use substitution to integrate $(r^{2}) \sqrt (r^{3} + 3) d r$ $(r^2) √(r^(3) +3)dr$ ?

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Answer 1 · 2018-03-25T22:47:42

$\int \frac{r^{2}}{\sqrt{r^{3} + 3}} d r = \frac{2}{3} \sqrt{r^{3} + 3} + C$ $intr^2/sqrt(r^3+3)dr=2/3sqrt(r^3+3)+C$

Explanation:

So, we want to determine

$\int \frac{r^{2}}{\sqrt{r^{3} + 3}} d r$ $intr^2/sqrt(r^3+3)dr$

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 $u = r^{3} + 3$ $u=r^3+3$ , as

$d u = 3 r^{2} d r$ $du=3r^2dr$

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

$\frac{1}{3} \int \frac{d u}{\sqrt{u}} = \frac{1}{3} \int u^{- \frac{1}{2}} d u$ $1/3int(du)/sqrt(u)=1/3intu^(-1/2)du$ (We've factored $\frac{1}{3}$ $1/3$ outside of the integral)

Integrate:

$(\frac{1}{3}) (2) \sqrt{u} + C$ $(1/3)(2)sqrt(u)+C$

Rewriting in terms of $x$ $x$ yields

$\int \frac{r^{2}}{\sqrt{r^{3} + 3}} d r = \frac{2}{3} \sqrt{r^{3} + 3} + C$ $intr^2/sqrt(r^3+3)dr=2/3sqrt(r^3+3)+C$

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How do you use substitution to integrate $(r^{2}) \sqrt (r^{3} + 3) d r$ $(r^2) √(r^(3) +3)dr$ ?

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Explanation:

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How do you use substitution to integrate $(r^{2}) \sqrt (r^{3} + 3) d r$ $(r^2) √(r^(3) +3)dr$ ?